First of all, note that
$$
\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})
\neq
\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ,
$$
since same functions in these products ($\psi_a$ and $\psi_b$) have different arguments ($\vec{r}_{1}$ and $\vec{r}_{2}$) to the left and to the right of the inequality sign.
Now, to the question, why do we have a linear combination. For a system of two indistinguishable particles (say, electrons) the obvious requirement is that there should be no observable difference between the system in state $\psi(1, 2)$ and the system in state $\psi(2, 1)$,
\begin{equation}
|\psi(1, 2)|^{2} = |\psi(2, 1)|^{2} \, ,
\end{equation}
where $1$ and $2$ stand for coordinates of particles.
This implies that
\begin{equation*}
\psi(1, 2) = \pm \psi(2, 1) \, .
\end{equation*}
And in order to satisfy this equality in what is often called the orbital approximation, where the state of a many electron system $\psi$ is represented as a product of a single-particle states $\psi_i$, the wave function $\psi(1, 2)$ should be actually written as not just a product of $\psi_1$ and $\psi_2$ (none of which, neither $\psi_1(1) \psi_2(2)$ nor $\psi_2(1) \psi_1(2)$, is symmetric or antisymmetric), but rather as the following linear combination of these products,
\begin{equation*}
\psi(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) \pm \psi_1(2) \psi_2(1) \Big) \, ,
\end{equation*}
where $1 / \sqrt{2}$ is the normalization factor.
The choice of the sign in the expressions above is not arbitrary, but rather determined by the particles spin $s$, namely,
\begin{equation*}
\psi(1, 2) = (-1)^{2s} \psi(2, 1).
\end{equation*}
Particles with integer spins, called bosons (photons, alpha particles), are described by a symmetric wave function
\begin{equation*}
\psi_{\mathrm{s}}(1, 2)
=
\frac{1}{\sqrt{2}}
\Big(
\psi_1(1) \psi_2(2)
+
\psi_1(2) \psi_2(1)
\Big) \, ,
\end{equation*}
while the particles with half-integer spins, called fermions (electrons, protons, neutrons), are described by the anti-symmetric wave function
\begin{equation*}
\psi_{\mathrm{a}}(1, 2)
=
\frac{1}{\sqrt{2}}
\Big(
\psi_1(1) \psi_2(2)
-
\psi_1(2) \psi_2(1)
\Big) \, .
\end{equation*}
