To go from the Hartree_Fock
$f\psi{_i}=\epsilon\psi{_i}$
to the Roothaan equation
$FC=SC\epsilon$.
equation we expand the orbitals as
$\psi{_i}=\sum C_{\mu i}\phi_i$
But for Helium atom we have just one orbital function so the coefficient $C_{\mu i}$ should be a vector not a matrix. My question is what is the interpretation of the matrix coefficient $C$ in the Roothaan equation $FC=SC\epsilon$ for the helium atom since we've got just one spatial orbital function?
The $\mathbf{C}$ matrix will always be $K\times K$, where $K$ is the size of the finite basis set you are using. That is to say, with a basis set of size $K$, you will produce exactly $K$ MOs. This means you will never wind up with $\mathbf{C}$ being a vector. In the example you linked, they use a basis set of size 2, so they get 2 MOs for helium and thus the $\mathbf{C}$ matrix is $2\times 2$. While helium has only one occupied spatial orbital, it will have $K-1$ virtual orbitals depending the size of the basis set you use to solve the restricted Roothaan-Hall equations.
