Tor and flat base change

It should be part of the hypotheses that $S$ is a flat $R$-module. I doubt the theorem is true for general ring extensions.

I suspect that what Rotman means by "flat on either side" is that it is flat on both sides.

This is true in general (assuming, as I'm sure was meant, that $B$ is flat "on either side" means on both sides), and I don't think you need the rings to be commutative. If $P_* \to A$ is a free resolution, then $P_* \otimes_R B$ has homology groups $Tor^R(A,B)$, and so it's also exact above degree zero. It's then a resolution too -- not by free or projective objects, but (since $P_*$ was free) by direct sums of copies of $B$. This makes $P_* \otimes_R B$ into a resolution of $A \otimes_R B$ by flat right $S$-modules, and that's good enough to compute Tor.

This makes the homology of the complex $(P_* \otimes_R B) \otimes_S C$ into something that computes $Tor_S(A \otimes_R B, C)$.

On the other hand, it is isomorphic to the complex $P_* \otimes_R (B \otimes_S C)$, which more directly is something that computes $Tor_R(A,B \otimes_S C)$.

(This feels a little ad-hoc. One perspective on Tor is that you're allowed to take a resolution of one side, or the other side, or of both sides and tensor together the resolutions; this is how you usually show that Tor over a commutative ring is symmetric. For this problem, that means that you can take resolutions of $A$ and $C$ simultaneously, which from a certain perspective might clean this argument up.)