Gopal Iyer's answer is only partially correct and doesn't contain the right interpretation of the citation in the question in my opinion. First I would like to shed some more light on Gopal Iyer's answer and then I will try to answer the actual question. From Gopal Iyer's answer:
Slater determinants are used to antisymmetrize the wave function so that fermion antisymmetry (Pauli principle) is obeyed. Since this is only relevant for electrons of the same spin, it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product.
As Antimon's comment correcly points out,
the fermionic wavefunction must be antisymmetric under the exchange of any two electrons.
Likewise the bosonic wave function must be symmetric under the exchange of any two particles. Here exchange of two particles means that all of the particle's parameters (i.e. the spatial parameter $\vec{x}_i$ and the spin parameter $s_i$) are exchanged. For fermions we have for example
$$\psi(\vec{x}_1,s_1,\vec{x}_2,s_2)=-\psi(\vec{x}_2,s_2,\vec{x}_1,s_1).$$
In many cases $\psi$ can be written as a product of a spatial function $\phi$ and a spin function $\chi$ and we have
$$\psi(\vec{x}_1,s_1,\vec{x}_2,s_2)=\phi(\vec{x}_1,\vec{x}_2)\chi(s_1,s_2),$$
$$\phi(\vec{x}_1,\vec{x}_2)\chi(s_1,s_2)=-\phi(\vec{x}_2,\vec{x}_1)\chi(s_2,s_1).$$
This means that if $\chi$ is a symmetric function, $\phi$ must be an antisymmetric function and vice versa. If both fermions occupy the same spin state $s$, then $\chi(s_1,s_2)=\delta_{ss_1}\delta_{ss_2}$ is a symmetric function and as a consequence $\phi$ must be an antisymmetric function. However, if the fermions don't occupy the same spin state, $\chi$ and $\phi$ can have any symmetry as long as the product is an antisymmetric function under exchange of the particles.
This is probably what Gopal Iyer's meant with:
Since this is only relevant for electrons of the same spin, [...].
However, the conclusion
it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product
is wrong. To correctly describe fermions, one has to use the full Slater determinant including the spin functions
$$\psi(\vec{x}_1,s_1,...,\vec{x}_N,s_N)=\frac{1}{\sqrt{N!}}\det\begin{pmatrix}
\psi_1(\vec{x}_1,s_1) & \dots & \psi_1(\vec{x}_N,s_N)\\
\vdots & \ddots & \vdots\\
\psi_N(\vec{x}_1,s_1) & \dots & \psi_N(\vec{x}_N,s_N)\\
\end{pmatrix}.$$
So, is the citation from the book in the question wrong then? The answer is no. The short answer is that the book doesn't claim that $\psi_{D}=\psi_{\alpha}\psi_{\beta}$ is the correct description of the fermions, nor does it claim that $\psi_{D}$ and the full antisymmetric $\psi$ are equivalent. The statement is:
In Monte Carlo, we may factor the determinant into electron spin components, i.e., $\psi_{D}=\psi_{\alpha}\psi_{\beta}$.
But what exactly does that mean? Why and how does it work? In order to get an understanding for that, I will first clarify why the Slater determinent is necessary in VMC at all.
In general, if we are looking for solutions (eigenstates) of a Schrödinger equation, we don't have to look for antisymmetric solutions. If we find an eigenstate that is not (fully) antisymmetric, we can always apply the antisymmetrizer or antisymmetrizing operator $\mathcal{A}$ in the end, provided that the found eigenstate is not symmetric under any odd parity permutation. If we use a Hartree product to construct our eigenstate, this can be ensured by using pairwise distinct single-particle functions $\psi_i$ (pairwise orthonormal spin-orbitals). The Slater determinant can then be written as
$$\psi(\vec{x}_1,s_1,...,\vec{x}_N,s_N)=\sqrt{N!}\mathcal{A}\psi_1(\vec{x}_1,s_1)...\psi_N(\vec{x}_N,s_N),$$
where $\psi_1(\vec{x}_1,s_1)...\psi_N(\vec{x}_N,s_N)$ is the Hartree product. However, the problem is that the Hartree product isn't an eigenstate of the Schrödinger equation in general. This is only the case in very simple problems (i.e. if the Schrödinger equation only contains single-particle operators, it is separable and the Hartree product of single-particle solutions is an eigenstate).
I will now show specifically for VMC, why using the simple Hartree product instead of the Slater determinant leads to a problem. In VMC the goal is to minimize the energy expectation value $\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}$ to get an approximation of the ground state. What would happen to the energy expectation value if we use the Hartree product instead of the fully antisymmetric $\psi$? If the result were the same, we could generally use the much simpler Hartree product instead to minimize the energy. Well, let's see. The Slater–Condon rules tell us
$$\langle\psi|\sum_{i=1}^{N}\hat{f}(\vec{x}_i,s_i)|\psi\rangle=\sum_{i=1}^{N}\langle\psi_{i}|\hat{f}|\psi_{i}\rangle,$$
$$\langle\psi|\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\hat{g}(\vec{x}_i,s_i,\vec{x}_j,s_j)|\psi\rangle=\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}(\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{i}\psi_{j}\rangle-\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{j}\psi_{i}\rangle).\ \ \ (1)$$
On the other hand, for the Hartree product we have simply
$$\langle\psi_1...\psi_N|\sum_{i=1}^{N}\hat{f}(\vec{x}_i,s_i)|\psi_1...\psi_N\rangle=\sum_{i=1}^{N}\langle\psi_{i}|\hat{f}|\psi_{i}\rangle,$$
$$\langle\psi_1...\psi_N|\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\hat{g}(\vec{x}_i,s_i,\vec{x}_j,s_j)|\psi_1...\psi_N\rangle=\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{i}\psi_{j}\rangle.\ \ \ (2)$$
The term $\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{j}\psi_{i}\rangle$ in equation $(1)$ isn't present in equation $(2)$, which means that the energy expectation value of the Hartree product and the fully antisymmetric $\psi$ are not generally equal
$$\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}\neq\frac{\langle\psi_1...\psi_N|\hat{H}|\psi_1...\psi_N\rangle}{\langle\psi_1...\psi_N|\psi_1...\psi_N\rangle}.$$
If the goal is to minimize the energy expectation value $\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}$, we can't minimize the energy expectation value with respect to the Hartree product instead because in general this could lead to a different minimum.
Side note: Although the Slater determinant is much harder to compute than the simple Hartree product, there are even some advantages. Due to the introduced (anti)symmetry of the Slater determinant, the variance in VMC is much smaller. Also the fact that the Slater determinant vanishes when two particles with the same spin function approach each other, makes it much easier to fulfill cusp conditions.
Above we have seen that the only difference between equation $(1)$ and $(2)$ is the term $\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{j}\psi_{i}\rangle$. However, this term vanishes if $\hat{g}$ is a spin-independent operator and $\psi_{i}$ and $\psi_{j}$ have different spin functions (i.e. $\alpha$, $\beta$). This gives rise to the assumption
$$\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}=\frac{\langle\psi_\alpha\psi_\beta|\hat{H}|\psi_\alpha\psi_\beta\rangle}{\langle\psi_\alpha\psi_\beta|\psi_\alpha\psi_\beta\rangle}.$$
I will now show this with an example Slater determinant
$$\psi(\vec{x}_1,s_1,...,\vec{x}_4,s_4)=\frac{1}{\sqrt{4!}}\det D,$$
$$D=\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_2)\alpha(s_2) & \phi_1(\vec{x}_3)\alpha(s_3) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_2)\alpha(s_2) & \phi_2(\vec{x}_3)\alpha(s_3) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_2)\beta(s_2) & \phi_3(\vec{x}_3)\beta(s_3) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_2)\beta(s_2) & \phi_4(\vec{x}_3)\beta(s_3) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix},$$
by using the hint in the citation in the question
all possible permutations of this term that are present in the full antisymmetric $\psi$ are operationally equivalent: each term simply corresponds to a relabeling of the electrons.
The same steps in the following calculation can be used to formulate a rigorous proof. $\hat{H}$ can be an arbitrary spin-independent operator in which all particles are indistinguishable.
So, how do we get the said factorization and the terms? We can do this by using the generalized Laplace expansion formula, which allows us to expanded along multiple rows or columns at once. We choose to expand along the rows with spin function $\alpha$ and define $I^\alpha=\{1,2\}$ to be the indices of those rows. The universe is $U=\{1,...,N\}=\{1,...,4\}$, an apostrophe $'$ denotes the complement of a set, vertical bars $|\cdot|$ denote the cardinality of a set and $D_{IJ}$ denotes the submatrix of $D$ with rows $I$ and columns $J$. Thus $I^\beta={I^\alpha}'=\{3,4\}$. Then the expansion is given by
$$\det D=\sum_{|J|=|I^\alpha|,J\subseteq U} (-1)^{\sum I^\alpha+\sum J}\det D_{I^\alpha J}\det D_{I^\beta J'}.$$
Using this formula we obtain
$$\det D=\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_2)\alpha(s_2)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_2)\alpha(s_2)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_3)\beta(s_3) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_3)\beta(s_3) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix}\\
-\det\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_3)\alpha(s_3)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_3)\alpha(s_3)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_2)\beta(s_2) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_2)\beta(s_2) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix}\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_2)\beta(s_2) & \phi_3(\vec{x}_3)\beta(s_3)\\
\phi_4(\vec{x}_2)\beta(s_2) & \phi_4(\vec{x}_3)\beta(s_3)\\
\end{pmatrix}\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_2)\alpha(s_2) & \phi_1(\vec{x}_3)\alpha(s_3)\\
\phi_2(\vec{x}_2)\alpha(s_2) & \phi_2(\vec{x}_3)\alpha(s_3)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix}\\
-\det\begin{pmatrix}
\phi_1(\vec{x}_2)\alpha(s_2) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_2)\alpha(s_2) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_3)\beta(s_3)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_3)\beta(s_3)\\
\end{pmatrix}\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_3)\alpha(s_3) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_3)\alpha(s_3) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_2)\beta(s_2)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_2)\beta(s_2)\\
\end{pmatrix}.$$
We immediately see that the first term is $\sqrt{2}\psi_\alpha\sqrt{2}\psi_\beta$ and all terms are equivalent up to relabeling of particles. From determinants we can factor out the spin functions per column
$$d:=\frac{1}{2}\det D=\\
+\psi_\alpha(\vec{x}_1,\vec{x}_2)\alpha(s_1)\alpha(s_2)\psi_\beta(\vec{x}_3,\vec{x}_4)\beta(s_3)\beta(s_4)\\
-\psi_\alpha(\vec{x}_1,\vec{x}_3)\alpha(s_1)\alpha(s_3)\psi_\beta(\vec{x}_2,\vec{x}_4)\beta(s_2)\beta(s_4)\\
+\psi_\alpha(\vec{x}_1,\vec{x}_4)\alpha(s_1)\alpha(s_4)\psi_\beta(\vec{x}_2,\vec{x}_3)\beta(s_2)\beta(s_3)\\
+\psi_\alpha(\vec{x}_2,\vec{x}_3)\alpha(s_2)\alpha(s_3)\psi_\beta(\vec{x}_1,\vec{x}_4)\beta(s_1)\beta(s_4)\\
-\psi_\alpha(\vec{x}_2,\vec{x}_4)\alpha(s_2)\alpha(s_4)\psi_\beta(\vec{x}_1,\vec{x}_3)\beta(s_1)\beta(s_3)\\
+\psi_\alpha(\vec{x}_3,\vec{x}_4)\alpha(s_3)\alpha(s_4)\psi_\beta(\vec{x}_1,\vec{x}_2)\beta(s_1)\beta(s_2).$$
Our goal is to calculate the expectation value
$$\langle\psi|\hat{H}|\psi\rangle=\left(\frac{2}{\sqrt{4!}}\right)^2\langle d|\hat{H}|d\rangle=\frac{1}{6}\langle d|\hat{H}|d\rangle.$$
Since $\hat{H}$ is a spin-independent operator, it doesn't effect the spin states when we apply it to $|d\rangle$. When $\langle d|$ is applied from the left side, we see that each cross term contains at least one $\langle\alpha(s_i)|\beta(s_i)\rangle=0$ or $\langle\beta(s_i)|\alpha(s_i)\rangle=0$. Thus only the direct terms remain with $\langle\alpha|\alpha\rangle=1$ and $\langle\beta|\beta\rangle=1$ and we get
$$\langle\psi|\hat{H}|\psi\rangle=\frac{1}{6}\\
(\langle\psi_\alpha(\vec{x}_1,\vec{x}_2)\psi_\beta(\vec{x}_3,\vec{x}_4)|\hat{H}|\psi_\alpha(\vec{x}_1,\vec{x}_2)\psi_\beta(\vec{x}_3,\vec{x}_4)\rangle\\
+\langle\psi_\alpha(\vec{x}_1,\vec{x}_3)\psi_\beta(\vec{x}_2,\vec{x}_4)|\hat{H}|\psi_\alpha(\vec{x}_1,\vec{x}_3)\psi_\beta(\vec{x}_2,\vec{x}_4)\rangle\\
+\langle\psi_\alpha(\vec{x}_1,\vec{x}_4)\psi_\beta(\vec{x}_2,\vec{x}_3)|\hat{H}|\psi_\alpha(\vec{x}_1,\vec{x}_4)\psi_\beta(\vec{x}_2,\vec{x}_3)\rangle\\
+\langle\psi_\alpha(\vec{x}_2,\vec{x}_3)\psi_\beta(\vec{x}_1,\vec{x}_4)|\hat{H}|\psi_\alpha(\vec{x}_2,\vec{x}_3)\psi_\beta(\vec{x}_1,\vec{x}_4)\rangle\\
+\langle\psi_\alpha(\vec{x}_2,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_3)|\hat{H}|\psi_\alpha(\vec{x}_2,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_3)\rangle\\
+\langle\psi_\alpha(\vec{x}_3,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_2)|\hat{H}|\psi_\alpha(\vec{x}_3,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_2)\rangle).$$
Since $\hat{H}$ doesn't distinguish particles, all six expectation values are equal to $\langle\psi_\alpha\psi_\beta|\hat{H}|\psi_\alpha\psi_\beta\rangle$ and we have
$$\langle\psi|\hat{H}|\psi\rangle=\langle\psi_\alpha\psi_\beta|\hat{H}|\psi_\alpha\psi_\beta\rangle.$$
So, what the book tells us is that we can either use the full antisymmetric $\psi$ or the factorized form $\psi_D=\psi_\alpha\psi_\beta$ in order to minimize the energy expectation value and we will get the same result. However, $\psi_D$ isn't the correct description of fermions. To get the correct fully antisymmetric description, one has to calculate $\psi=\frac{\mathcal{A}\psi_D}{\langle\mathcal{A}\psi_D|\mathcal{A}\psi_D\rangle}$.
