Why does flatness imply these Homs are isomorphic? I just encountered a piece of reasoning that I assume is standard but it's unfamiliar to me. Do you know a proof or a standard reference?
Setting: Suppose $R\rightarrow S$ is a map of commutative, unital rings, and $M,U$ are $R$-modules and $N$ an $S$-module. We have a natural map
$$ \operatorname{Hom}_R(U,M)\otimes N \rightarrow \operatorname{Hom}_S(U\otimes S, M\otimes N)$$
given by $f\otimes n \mapsto f\otimes (1\mapsto n)$, with $f\in \operatorname{Hom}_R(U,M), n\in N$. (All tensor products are taken over $R$.)

Question: if $N$ is flat as an $R$-module, is this map always an isomorphism. If so, why?

This comes up on p. 13 of Bruns and Herzog's book Cohen-Macaulay Rings from 1993. In this context there are several other assumptions in place that I was assuming are extraneous to this particular point, but I would be happy to know otherwise. They are: (1) $R,S$ are Noetherian local rings; (2) $M,N$ are finite modules over $R,S$ respectively; and (3) $U$ is the residue field of $R$.
 A: The RHS is equivalently $\text{Hom}_R(U, M \otimes_R N)$ so let's concentrate on that instead. Note that this means the question no longer has anything to do with $S$.
One way to interpret the question is that you're asking when $(-) \otimes_R N$ preserves a functor which behaves like a limit, namely $\text{Hom}_R(U, -)$ (to think of this functor as a limit pick a presentation of $U$). Now, tensoring with flat modules preserves finite limits, and it's not hard to show from here that the natural map is an isomorphism whenever $U$ is finitely generated and $N$ is flat. Arguing dually you also get an isomorphism whenever $N$ is finitely generated and $U$ is projective. 
For a counterexample to having an isomorphism in the requested level of generality, take $R = \mathbb{Z}, U = \mathbb{Q}, M = \mathbb{Z}, N = \mathbb{Q}$. Note here that neither $U$ nor $N$ is finitely generated. The problem is that although $N$ is flat, this only guarantees that $(-) \otimes N$ preserves finite limits, and $\text{Hom}(\mathbb{Q}, -)$ is an infinite limit. 
