So just thinking about Spin and not the helium atom for the moment,
$$S=S_1+S_2, \quad S^2=S_1^2+2S_1\cdot S_2+S_2^2, \quad S_z=S_{1z}+S_{2z}$$
We can also use that $S_+=S_x+iS_y$ and $S_-=S_x-iS_y$ that
$$S_1\cdot S_2=S_{1z}S_{2z}+\frac{1}{2}\left(S_{1+}S_{2-}+S_{1-}S_{2+}\right)$$
Anyway what will be the old spin basis will be common eigenvectors of the commuting observables $$\{S_1^2, S_2^2,S_{1z},S_{2z}\}$$
These could be notated e.g. $\left|\uparrow\uparrow\right\rangle=\left|\uparrow\right\rangle\otimes\left|\uparrow\right\rangle=\chi^+(1)\chi^+(2)$. In the new spin basis we take the vectors to be common eigenvectors of the commuting observables
$$\{S_1^2, S_2^2,S^2,S_{z}\}$$
We just change basis and as you have it in the new basis the eigenstates are
\begin{align}
\left|1,1\right\rangle&=\left|\uparrow\uparrow\right\rangle\\
\left|1,0\right\rangle&=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle+\left|\downarrow\uparrow\right\rangle\right)\\
\left|1,-1\right\rangle&=\left|\downarrow\downarrow\right\rangle\\
\left|0,0\right\rangle&=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle-\left|\downarrow\uparrow\right\rangle\right)\\
\end{align}
You can check using the definitions of $S_z$ and $S^2$ that theses are eigenvectors, e.g.
$$S_z\left|1,1\right\rangle= (S_{1z}+S_{2z})\left|\uparrow\uparrow\right\rangle=S_{1z}\left|\uparrow\uparrow\right\rangle+S_{2z}\left|\uparrow\uparrow\right\rangle$$
$$S_1\left|\uparrow\uparrow\right\rangle=\left(S_1\left|\uparrow\right\rangle\right)\otimes \left|\uparrow\right\rangle=\frac{\hbar}{2}\left|\uparrow\uparrow\right\rangle$$
$$S_2\left|\uparrow\uparrow\right\rangle=\left|\uparrow\right\rangle\otimes \left(S_2\left|\uparrow\right\rangle\right)=\frac{\hbar}{2}\left|\uparrow\uparrow\right\rangle$$
So,
$$S_z\left|1,1\right\rangle= \hbar\left|\uparrow\uparrow\right\rangle=\hbar \left|1,1\right\rangle$$
Another example is $S^2$,
\begin{align}
S^2 \left|1,1\right\rangle&=\left(S_1^2+S_2^2 +2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}\right)\left|\uparrow,\uparrow\right\rangle\\
&=\left(\frac{3}{4}\hbar^2+\frac{3}{4}\hbar^2 +2\frac{\hbar}{2}\frac{\hbar}{2}+0+0\right)\left|\uparrow,\uparrow\right\rangle\\
&=\left(\frac{8}{4}\hbar^2\right)\left|\uparrow,\uparrow\right\rangle\\
&=2\hbar^2\left|1,1\right\rangle\\
&=(s)(s+1)\hbar^2\left|1,1\right\rangle\quad \text{where } s=1.
\end{align}
You get the second line similar to the last example and using how the single spin operators act on a single spin, in short:
\begin{align}
S^2\left|\uparrow\right\rangle&=\frac{3}{4}\hbar^2\left|\uparrow\right\rangle\\
S_z\left|\uparrow\right\rangle&=\frac{\hbar}{2}\left|\uparrow\right\rangle\\
S_+\left|\uparrow\right\rangle&=0\\
S_-\left|\downarrow\right\rangle&=0
\end{align}
Anyway, back to the helium atom, depending on the levels of approximation (order in perturbation theory etc) you might try write say the ground state as
$$\psi=\phi_{\text{orbital}}\cdot\chi_{\text{spin}}$$
or with kets,
$$\left|\psi\right\rangle=\left|\phi_{\text{orbital}}\right\rangle\otimes\left|\chi_{\text{spin}}\right\rangle$$
Written as such the spin operator on acts on the $\chi$ part of the wave function, so if this is a spin singlet or triplet then it is and eigenvalue of $S^2$ and $S_z$, e.g. if $\left|\chi\right\rangle=\left|s,m\right\rangle$ then
\begin{align}
S^2\left|\psi\right\rangle&=S^2\left(\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\right)\\
&=\left|\phi_{\text{orbital}}\right\rangle\otimes\left(S^2\left|s,m\right\rangle\right)\\
&=s(s+1)\hbar^2\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\\
\end{align}
and
\begin{align}
S_z\left|\psi\right\rangle&=S_z\left(\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\right)\\
&=\left|\phi_{\text{orbital}}\right\rangle\otimes\left(S_z\left|s,m\right\rangle\right)\\
&=m\hbar\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\\
\end{align}
I don't know all of the details of the exact calculations of Helium ground and excited states, or the energies however.
Best Answer
For two electrons your are right! The absolute must is for the total state to be anti-symmetric. Technically, this means that the total wave-function must belong to a one-dimensional representation of the permutation group such that each permutation is represented by either +1 or -1 depending on it's parity.
You can ask about permutational symmetry separately for the spin part and the orbital part. If there number of electrons $n>2$, then the orbital part and the spin part may belong to these may belong to more complicated (e.g., more than one-dimensional) representations of the permutations group, those get classified by Young tableux..There are rules how to combine two "conjugate"(not sure if this is mathematically correct term) representations to construct the total wave-function anti-symmetric under any odd permutation. For $n=2$ these rules reduce to a simple product as you quote. For more on this you can consult Landau and Lifshitz, Quantum Mechanics (The Course of Theoretical Physics vol.III).
There is also a systematic connection between the representations of the permutation group for the spin part and the $SU(2)$ of the spin. Unfortuntally, I can't find the refernce which I have learn this from (hopefully others can help).