For some context, I am trying to prove that extension and restriction of scalars are an adjoint pair. Formally, given a commutative ring map $f: A \rightarrow B$ and $M$ an $A$-module, $N$ a $B$-module I need to show that the $B$-module morphisms from $B \otimes_A M$ to $N$ are the same as the $A$-module morphisms from $M$ to $N$.
Morally, I see that a $B$-module morphism from $B \otimes_A M$ is given by how it acts on the "generators" of the form $1 \otimes_A m$ by $$\varphi(b \otimes m) = \varphi(b \cdot (1 \otimes m)) = b \cdot \varphi(1 \otimes m).$$ I think that the submodule made of elements of this form is isomorphic to $M$ (so of course the maps would be the same), but I'm not sure what to call it. It's not $0 \otimes_A M$ (because this is isomorphic to the $0$ module), but I'm not sure why it'd be justified to call it $A \otimes_A M$ (which I know is isomorphic to $M$, i.e. what I want).
EDIT: I know the title uses different variables, but I wanted to make the crux of my question clear to the title-skimmer while providing all the context to the actual question-reader. Please let me know if this is not the correct thing to do.
I think you have a small error in the bijection you want to show. You want to show there is a bijection between $A$-module morphisms $M \to N$ (where $N$ is viewed as an $A$-module through $f$) and $B$-module morphisms $B \otimes_A M \to N$. (Instead of $B$-module morphisms $M \to N$ and $A$-module morphisms $B \otimes_A M \to N$.
Then, given an $A$-module morphism $\varphi : M \to N$, you can define $\psi : B \otimes_A M \to N$ by sending $b \otimes m$ to $b\varphi(m)$, and you can check this gives a well-defined $B$-morphism.
To go the other way, given a $B$-module morphism $\psi : B \otimes_A M \to N$, you can define the $A$-module morphism $\varphi : M \to N$ by sending $m$ to $\psi(1 \otimes m)$, and again check that this is an $A$-module morphism.