Let $A \to B$ be a homomorphism of commutative rings and $M,N$ be two $A$-modules. Consider the natural homomorphism of $B$-modules
$\alpha_{M,N} : \mathrm{Hom}_A(M,N) \otimes_A B \to \mathrm{Hom}_B(M \otimes_A B,N \otimes_A B)$
defined by $\alpha_{M,N}(f \otimes b)(m \otimes b') := f(m) \otimes bb'$. Consider the two statements:
(1) $\alpha_{M,N}$ is an isomorphism for all $M,N$ whenever $M$ is finitely presented.
(2) $A \to B$ is flat.
Then it is easy to see that (2) $\implies$ (1). Namely, $\alpha_{-,N}$ is a natural transformation between finitely continuous functors and $\alpha_{A,N}$ is obviously an isomorphism, thus $\alpha_{M,N}$ is an isomorphism whenever $M$ is finitely presented.
Question. Does also (1) $\implies$ (2) hold? If this is false for trivial reasons, can we strenghten (1) a little bit so that it becomes true?
For example, one might try:
(1') For all $N$ the locus where $\alpha_{-,N}$ is an isomorphism is closed under finite colimits, or equivalently, under cokernels.
Then we still have (2) $\implies$ (1') and one might ask (1') $\implies$ (2).