Given a ring morphism from $B$ to $A$, we can regard an $A$-module $M$ as a $B$-module. Then how can I prove the functor $._B:\operatorname{Mod}_A \to \operatorname{Mod}_B$ is right adjoint to $.\otimes_BA$, i.e., $\operatorname{Hom}_A(N\otimes A, M)\cong \operatorname{Hom}_B(N,M_B)$?
I guess this is some kind of standard result, so where can I find some reference?
It's the standard adjunction with the Hom functor, namely $$ M_B\cong\operatorname{Hom}_A(A,M) $$ regarding $A$ as a $B$-module by means of the given homomorphism.
Another way to see this is by building the natural bijection by hand. If $f\colon N\otimes_BA\to M_B$ is a homomorphism of $B$-modules, define $\tilde{f}\colon N\to M$ by $\tilde{f}(x)=f(x\otimes 1)$. Prove this is a homomorphism of $B$-modules.
The inverse map sends $g\colon N\to M_B$ to $\hat{g}$ defined via the $B$-balanced map $\bar{g}\colon N\times A\to M$ by $$ \bar{g}(x,a)=g(x)a $$