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.
The total wave function needs to be antisymmetric under particle interchange. Since each electron is in the same 1-particle ground state, $E_0(x)$, the spatial wave function will be symmetric under interchange; hence, the spin wave function must be antisymmetric.
The 2-particle wave function is:
$$ E_{0,0}(x_1, x_2) = E_0(x_1)E_0(x_2) = E_0(x_2)E_0(x_1) = E_0(x_2, x_1).$$
The rules regarding addition of angular momentum are well documented. The antisymmetric ground state will have $S=0$, and of course $S_z=0$:
$$\Xi_{1, 2} = \frac 1{\sqrt{2}}(\uparrow_1\downarrow_2-\downarrow_1\uparrow_2),$$
where the subscripts label the particle index (and the arrows indicate the z-component). Note that:
$$\Xi_{2, 1} = \frac 1{\sqrt{2}}(\uparrow_2\downarrow_1-\downarrow_2\uparrow_1) = -\frac 1{\sqrt{2}}(\uparrow_1\downarrow_2-\downarrow_1\uparrow_2)=-\Xi_{1, 2}$$
so that the spin state is indeed antisymmetric.
The total wave function is their product:
$$ \psi_{1, 2} = E_0(x_1)E_0(x_2)\Xi_{1, 2}.$$
Note that the statement "the electrons have different spin" is misleading (I would even say "classical"): they have the same spin: $J = \hbar\sqrt{j(j+1)} = \sqrt{3/2}\hbar$. They even have the same projection onto the $z-$axis: $\pm\hbar/2$--it's just that their combination is antisymmetric under interchange.
Finally: nowhere did I need to refer to the quantitative solution of the square well.
Best Answer
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 $2E_1$.