I'm confused about a number of things concerning two-electron systems and spin. Here is (perhaps too much) exposition, skip to "the problem" if you want:
Consider the helium atom in the simplified picture where we ignore electron-electron repulsion, and let $\psi_{nlm}$ be the usual space-wavefunctions, and let $\chi_{\uparrow}$, $\chi_{\downarrow}$ denote the simultaneous eigenfunctions of $\mathbf{S}^2$ and $S_z$. Then write the one-electron wavefunctions as $$ \phi(i) = \psi(\mathbf{r}_i) \chi(s_i), \qquad i \in \{1,2\}$$ where $i$ is short for the space and spin coordinates $(\mathbf{r}_i, s_i)$. Given two one-electron wavefunctions $\phi_1$, $\phi_2$ we can always form a corresponding antisymmetric two-electron wavefunction with the Slater determinant: $$\phi(1,2) = \frac{1}{\sqrt{2}} \det \begin{bmatrix} \phi_1(1) & \phi_1(2) \\ \phi_2(1) & \phi_2(2) \end{bmatrix}.$$
Now, I'm having trouble constructing the eigenstates of helium (in this simplified picture). Following Griffiths, we assume one electron is in "the ground state" (i.e. has space-wavefunction $\psi_{100}$) while the other electron is in some "state" $\psi_{nlm}$.
In the case where both electrons have the space-wavefunction $\psi_{100}$ then we let $$\phi_1 = \psi_{100} \chi_{\uparrow}, \qquad \phi_2 = \psi_{100} \chi_{\downarrow}$$ and using the Slater determinant, we get $$\phi(1,2) = \psi_{100}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2) \left[\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) - \chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right].$$ We recognise the spin part as the "singlet", with total spin $0$. We also notice that this is the only possible wavefunction (up to overall phase) where both electrons are in the "$\psi_{100}$ state". I feel like I understand this case but I included it to make sure we're on the same page.
The problem (tl;dr): Consider now the case where one electron is in $\psi_{100}$ and the other is in some other "state", e.g. $\psi_{200}$. Griffiths tells me that we find possible two-electron states by constructing the symmetric and antisymmetric space-wavefunctions $$ \psi_{\text{S}}(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}} \left(\psi_{100}(\mathbf{r}_1)\psi_{200}(\mathbf{r}_2) + \psi_{200}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2) \right),$$ $$ \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}} \left(\psi_{100}(\mathbf{r}_1)\psi_{200}(\mathbf{r}_2) - \psi_{200}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2) \right)$$ and pairing them with an antisymmetric and antisymmetric spin-functions respectively: $$\phi(1,2) = \psi_{\text{S}}(\mathbf{r}_1, \mathbf{r}_2)\left[\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) - \chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right], \qquad \text{(singlet)}$$ $$\phi(1,2) = \begin{cases} \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2)\chi_{\uparrow}(s_1)\chi_{\uparrow}(s_2) & \\ \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2)\left[\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) + \chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right] & \text{(triplet)} \\ \psi_{\text{A}}(\mathbf{r}_1, \mathbf{r}_2)\chi_{\downarrow}(s_1)\chi_{\downarrow}(s_2) & \end{cases}$$
I can see why these are valid states, but it seems to me that these are not all the possible states. For example, letting
$$\phi_1 = \psi_{100} \chi_{\uparrow}, \qquad \phi_2 = \psi_{200} \chi_{\downarrow}$$
then using the Slater determinant, we get another antisymmetric wavefunction with total spin $0$:
$$\phi(1,2) = \frac{1}{\sqrt{2}} \left[\psi_{100}(\mathbf{r}_1)\psi_{200}(\mathbf{r}_2)\chi_{\uparrow}(s_1)\chi_{\downarrow}(s_2) - \psi_{200}(\mathbf{r}_1)\psi_{100}(\mathbf{r}_2)\chi_{\downarrow}(s_1)\chi_{\uparrow}(s_2) \right] .$$
How do these two approaches fit together? Is it possible to form this last wavefunction by taking linear combinations of the singlet/triplet wavefunctions? Can you form the singlet/triplet wavefunctions with slater determinants?
Also, is there reason to prefer one approach to the other?
