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 $2E_1$.
Best Answer
In the ground state, both electrons are in the state with the lowest value of "n". E.g., in the case of an infinite potential well, the lowest quantum number is n=1. In this case, both electrons have n=1, but one electron is spin up and one electron is spin down, because of exclusion.
The first excited state is one in which one of the electrons has n=1 (lowest single particle level) and one of the electrons has n=2 (first excited single particle level). In this case, either spin is okay for either electron.
So there are four states: (n=1,up; n=2,up), (n=1,up; n=2, down), (n=1,down; n=2, up), (n=1,down; n=2,down).
And, No, the wave function is not just the superposition for each one. The wave function is a Slater Determinant of the single-particle wavefunctions.
For example, in the case of the ground state, the spatial part of the wavefunction is symmetric $$ \sin(x_1\pi/L)\sin(x_2\pi/L) $$ and the spin part of the wavefunction is anti-symmetric $$ |\uparrow\downarrow>-|\downarrow\uparrow>\;. $$
You can work out the four excited states similarly.