The particle of mass $m$ in the box of length $L$ in 1D is solved by wavefunctions
$$
\begin{align}
\psi_{n\alpha}&=A\sin (k_n x) e^{-\omega_n t}|\alpha \rangle\;, \\
k_n&=\frac{n\pi}{L}\;,\\
E_n&=\hbar \omega_n\;,\\
\omega_n&=\frac{\pi h n^2}{4L^2m}\;.
\end{align}
$$
Here, $|\alpha \rangle$ represents the spin state.
The global fermionic wavefunction for two particles is constructed from all pairs by anti-symmetrization, as
$$
\Psi_{n\alpha m\beta}(x_1,x_2,t)=\psi_{n\alpha}(x_1,t)\psi_{m\beta}(x_2,t) - \psi_{m\beta}(x_1,t)\psi_{n\alpha}(x_2,t)\;.
$$
Energy of state $\Psi_{n\alpha m\beta}(x_1,x_2,t)$ can be calculated as
$$
(H_1+H_2)\Psi_{n\alpha m\beta}(x_1,x_2,t)=(E_n+E_m)\Psi_{n\alpha m\beta}(x_1,x_2,t)\;,
$$
since each of the one-particle Hamiltonians acts on the respective one-particle wavefunction $\psi_{n\alpha}(x_1,t)$, which yields its eigenenergy $E_n$.
For identical spins, we are interested only in solutions for which $\alpha=\uparrow$ and $\beta=\uparrow$. The ground state is the lowest lying energy state of the system. In this case, it would correspond to $\Psi_{1\uparrow 1\uparrow}$, but this function is identically zero. Then next two lowest lying states are $\Psi_{1\uparrow 2\uparrow}$ and $\Psi_{2\uparrow 1\uparrow}$. Thanks to the antisymmetrization, $\Psi_{1\uparrow 2\uparrow} = -\Psi_{2\uparrow 1\uparrow}$ and it represents the ground state of the system with energy $E_1+E_2$. For opposite spins, we choose $\alpha=\uparrow$ and $\beta=\downarrow$. Here, the lowest lying energy state is $\Psi_{1\uparrow 1\downarrow}$ and it has energy $2E_1$.
First, observe that the Slater determinant you have written down is the linear combination of the singlet state and the z-spin-0 state of the triplet. Vice versa, you can produce the singlet and triplet states as linear combinations of Slater determinants.
Whether you prefer the Slater determinant or the singlet/triplet formalism for writing down your two-fermion states depends on your personal preference and on the specific application you have in mind.
Best Answer
A better notation would be $$ \vert +\rangle_1\vert +\rangle_2\, ,\quad \vert +\rangle_1\vert -\rangle_2\, ,\quad \vert -\rangle_1\vert +\rangle_2\, ,\quad \vert -\rangle_1\vert -\rangle_2 $$ with the particle label explicitly attached. Note that the labelling matters in the sense that $\vert +\rangle_1\vert -\rangle_2$ is not the same as $\vert -\rangle_1\vert +\rangle_2$ since the former means the first particle has spin up and the second has spin down, while the latter means the first has spin down and the second has spin up.
The symmetry/antisymmetry comes from permuting the particle labels $1$ and $2$. Permuting these labels in the $S=1$ states transforms the states back into themselves, v.g. $$ P_{12}\vert +\rangle_1\vert +\rangle_2=\vert +\rangle_2\vert +\rangle_1= \vert +\rangle\vert +\rangle_2 $$ but permuting the labels in the $S=0$ state gives the negative of the original state: $$ P_{12}\left(\vert +\rangle_1\vert -\rangle_2-\vert -\rangle_1\vert +\rangle_2\right)= \left(\vert +\rangle_2\vert -\rangle_1-\vert -\rangle_2\vert +\rangle_1\right)=-\left(\vert +\rangle_1\vert -\rangle_2-\vert -\rangle_1\vert +\rangle_2\right)= $$ after reordering so that kets for particle 1 occurs before those of particle 2.
Edit.
The state $\vert S,m\rangle$ with \begin{align} \vert 1,1\rangle &=\vert +\rangle_1\vert +\rangle_2\, ,\\ \vert 1,0\rangle&=\frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2+ \vert -\rangle_1\vert +\rangle_2\right)\\ \vert 1,-1\rangle&=\vert -\rangle_1\vert -\rangle_2\\ \vert 0,0\rangle&=\frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2- \vert -\rangle_1\vert +\rangle_2\right) \end{align} are just rearrangements of the original state so they are eigenstates of $S^2$ with eigenvalue $S(S+1)$, and eigenstates of $S_z$ with eigenvalue $m$. Thus, for instance, $\vert 1,0\rangle$ is eigenstate of $S^2$ with eigenvalue $2$ and eigenstate of $S_z$ with eigenvalue $0$.
If you start a space spanned by $4$ states, you cannot rearrange the basis so there are fewer than $4$ states after the rearrangement. Thus, in addition to the $\vert 1,m\rangle$ state, you also need the $\vert 00\rangle$ state.
Note that all $\vert 1,m\rangle$ states are symmetric under permutation of the particle labels, where $\vert 00\rangle$ is antisymmetric. In addition, you can reach any state in the $\vert 1,m\rangle$ set by raising or lowering with $S_\pm$, but the action of $S_\pm$ will not take you from an $\vert 1m\rangle$ to the $\vert 00\rangle$ state: in general, $S_\pm$ and $S_z$ cannot change the value of $S$.
As a final note, the operators $S_\pm$ and $S_z$ are defined through $$ S_z= S_{1,z}+S_{2,z},\qquad S_+= S_{1,+}+S_{2,+}\qquad S_-= S_{1,-}+S_{2,-} $$ where $S_{k,z}, S_{k,+}$ and $S_{k,-}$ act only on the ket for particle $k$.