The total wave function needs to be antisymmetric under particle interchange, e.g.:
$$\Psi(x_1, x_2) = - \Psi(x_2, x_1) $$
where the total wave function is factored into space, spin, and maybe color, and so on:
$$\Psi(x_1, x_2) = \psi(x_1, x_2)\chi_{1, 2}C_{1, 2} $$
where $C=1$ for colorless electrons.
As you point out, the spin can be $S=0$ or $S=1$. Spin zero is antisymmetric:
$$ \chi_{1, 2}=\frac 1{\sqrt 2}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle) = -\chi_{2, 1} = -[\frac 1{\sqrt 2}(|\downarrow\uparrow\rangle-|\uparrow\downarrow\rangle)]$$
while spin one are the three symmetric combinations, such as:
$$ \chi_{1,2}=|\uparrow\uparrow\rangle=+\chi_{2,1}=+[|\uparrow\uparrow\rangle]$$
For the total wave function to be antisymmetric, the spatial part of the wave function needs to have opposite symmetry, so that if the ground state is spin zero:
$$ \psi(x_1, x_2) = \psi(x_2, x_1) $$
so that:
$$ \Psi(x_1, x_2) = \psi(x_1, x_2)\chi_{1, 2} = [+\psi(x_2, x_1)][-\chi_{2, 1}] = -\Psi(x_2, x_1)$$
Particle interchange means interchanging all coordinates: space, spin, color, flavor, and whatever else may be relevant.
Now the comment that the total spin of the system is 0 or 1 and hence is symmetric is interesting. Yes, the 2 particles form a composite boson, but there is only one boson, and hence, nothing to interchange.
The best example of this is $^4$He. It is obviously a boson, as it forms a condensate--superfluid helium, and indeed, the superfluid wave function is symmetric under interchange of any two helium atoms. Nevertheless, the 2 electrons, 2 protons, and 2 neutrons in each helium atom are antisymmetric under interchange.
Best Answer
I'm using the standard notation throughout the whole answer. The problem of adding angular momenta is essentially a change of basis, from one that diagonalizes $(S_1^2,S_2^2,S_{1z},S_{2z})$ to one that diagonalizes $(S^2,S_z,S_1^2,S_2^2).$ If you work out the problem which is given in many texts you will find the following transformation.
$$|s=1m=1,s_1=1/2 s_2=1/2\rangle =|++\rangle$$ $$|s=1m=0,s_1=1/2 s_2=1/2\rangle =2^{-1/2}[|+-\rangle+|-+\rangle]$$ $$|s=1m=-1,s_1=1/2 s_2=1/2\rangle =|--\rangle$$ $$|s=0m=0,s_1=1/2 s_2=1/2\rangle =2^{-1/2}[|+-\rangle-|-+\rangle]$$
The allowed values for total spin are $s=1$ and $0$,while the allowed values of $s_z$ are $\hbar,0$ and $-\hbar$.
For a system of two spins 1/2 particles the wavefunction have the following possible forms
$$\Psi = \left\{ \begin{array}{l} \psi_a\chi_s \\ \psi_s\chi_a \end{array} \right.$$ subscript $s$ and is to denote symmetric and anti-symmetric.
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.
$$\Psi_{n\alpha m\beta}(x_1,x_2,t)=\psi_{n\alpha}(x_1,t)\psi_{m\beta}(x_2,t) - \psi_{m\beta}(x_1,t)\psi_{n\alpha}(x_2,t)$$
The 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 $E_n$.
First consider state 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 $E_1+E_2$. So the first two lowest energies are $$E^0=E^1=5E_0$$
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 $2E_0$.
You may wonder because this doesn't match with the answer in the textbook, So the only thing I can conclude that there is a mistake in the problem or in solution. I hope this will help you. Best wishes!