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!
Your intuition is correct.
Consider a system with a fixed number $N$ of nonrelativistic particles, all fermions. Ignoring spin for simplicity, the wavefunction of such a system is a function of $N$ points in space:
$$
\newcommand{\bfx}{\mathbf{x}}
\psi(\bfx_1,\bfx_2,...,\bfx_N).
\tag{1}
$$
If the $j$th and $k$th particles are the same species, then the wavefunction must satisfy
$$
\newcommand{\bfx}{\mathbf{x}}
\psi(\bfx_{\pi(1)},\bfx_{\pi(2)},...,\bfx_{\pi(N)})
=
-
\psi(\bfx_1,\bfx_2,...,\bfx_N)
\tag{2}
$$
for the permutation $\pi$ that exchanges $j\leftrightarrow k$ and leaves the other points unchanged. If the $j$th and $k$th particles are not the same species, then no such (anti)symmetry is required. In particular, if we have $5$ particles ($N=5$) all of different species, then the wavefunction doesn't need to have any (anti)symmetry at all. The fact that the particles are all fermions is irrelevant in that case. Nothing would change if they were bosons — in the strictly nonrelativistic and spinless model that we're using here for simplicity. (In a relativistic model, we can't ignore spin, and the number of particles is usually ill-defined, but I won't go into those complications here.)
We can consider all values of $N$ simultaneously using the formalism of creation/annihilation operators. Still ignoring spin for simplicity, we can describe a system of strictly nonrelativistic fermions using $K$ different creation operators $a_k^\dagger(\bfx)$ for each $\bfx$, with $k\in\{1,2,...,K\}$, where $K$ is the number of different species. If $|0\rangle$ is the state with no particles, then
$$
\int_{\bfx,\bfx'} \psi(\bfx,\bfx')
a_1^\dagger(\bfx)a_1^\dagger(\bfx')|0\rangle
\tag{3}
$$
is an example of a state with two particles of the same species, and
$$
\int_{\bfx,\bfx'} \psi(\bfx,\bfx')
a_1^\dagger(\bfx)a_2^\dagger(\bfx')|0\rangle
\tag{4}
$$
is an example of a state with two particles of different species. The assertion that the particles are all fermions can be expressed mathematically by the requirement that all of the creation operators anticommute with each other:
$$
a_j^\dagger(\bfx)a_k^\dagger(\bfx')
=
-a_k^\dagger(\bfx')a_j^\dagger(\bfx).
\tag{5}
$$
For particles of the same species ($j=k$), this immediately implies the Pauli exclusion principle: only the antisymmetric part of $\psi$ matters in (3), because the minus sign in (5) eliminates any contribution from the symmetric part. This follows from the fact that in (3), exchanging the subscripts is the same as exchanging the points. But for particles of different species ($j\neq k$), that's no longer true, and the minus sign in (5) has no consequence in thiscase. In fact, the minus sign in (5) can be eliminated when $j\neq k$ using a Klein transform.
Best Answer
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.