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.