Let's consider two electrons. The spin operator of the system is (omitting vector symbols for simplicity):
$S=S_1\otimes\mathbb I_2+\mathbb I_1 \otimes S_2$
I want to show that an eigenbasis of {$S^2,S_z$} is given by {$\chi_u, \chi_+, \chi_-,\chi_d$} where
$$\chi_u=|uu\rangle$$ $$\chi_d=|dd\rangle$$ $$\chi_+=\frac{1}{\sqrt2}|ud+du\rangle$$ $$\chi_-=\frac{1}{\sqrt2}|ud-du\rangle$$ here $|u\rangle$ is an eigenvector of $S_i^2$ and of $S_{z,i}$ of eigenvalues $\frac{1}{2}$ and $\frac{1}{2}$ and $|d\rangle$ is an eigenvector of $S_i^2$ and of $S_{z,i}$ of eigenvalues $\frac{1}{2}$ and $-\frac{1}{2}$ respectively and for $i=1,2$.
I can figure out how $S_z$ works on this system of two particles, but I'm not quite sure what $S^2$ is here. My euristhic guess (I'm just doing a dot product) is that we could have:
$S^2=S_1^2\otimes\mathbb I_2+\mathbb I_1 \otimes S_2^2+2S_1\otimes S_2$
but:
I don't know if this is true
If 1. is true, then the $2S_1\otimes S_2$ is very annoying and I don't really know how to deal with this and thus how to find the eigenbasis of {$S^2,S_z$}