A map of $A$-modules is an isomorphism for any $M$,$N$ iff $r: A \rightarrow B$ is an epimorphism

Question

I'm trying to understand the accepted answer in this question.

Context: Let $f: R \rightarrow S$ be a morphism of commutative rings. Let $M,N$ be $S$-modules. Then there is a natural morphism $\phi : M \otimes_R N \rightarrow M \otimes_S N$ of $R$-modules given by sending $m \otimes n \mapsto m \otimes n$. We want to show that $\phi$ is an isomorphism for all $M$ and $N$ if and only if $f$ is an epimorphism.

I've narrowed my problems down to the following phrases in the accepted answer:

A) The equivalence "$S \otimes_R S \rightarrow S \otimes_S S \cong S$ is an isomorphism if and only if $f:R \rightarrow S$ is an epimorphism";

B) The condition '$m: S \otimes_R S \rightarrow S$ is an isomorphism' is "sufficient because if $f$ is an isomorphism then it induces a natural transformation which is also an isomorphism";

C) "Formally, there's a functor (even a monoidal functor) from the category of $(S,S)$-bimodules and bimodule homomorphisms to the category of functors $\text{Mod}(S) \times \text{Mod}(S) \rightarrow \text{Ab}$ and natural transformations, and it sends isomorphisms to isomorphisms".

My specific questions:

1. I'd like a proof of (A).
   Context: The question seems to imply that we want to apply the fact about restriction of scalars being fully faithful when $f$ is an epimorphism (proof here), but I don't see how $S \otimes_R S$ is obtained by restricting scalars from some $S$-module.

2. I'd like an explanation of (B).
   In particular, what functors is the natural transformation between, why is it an isomorphism, and why do we need $f$ to be an isomorphism?

3. I'd like an explanation of why (B) means that $m$ being an isomorphism is sufficient.
   Context: I think it is clear that $m$ being an isomorphism is sufficient, because we just tensor with $\text{id}$ on both sides which gives us an isomorphism. Is this all that is being abstractified in this natural transformation or is something else going on here? Again, why do we require $f$ to be an isomorphism?

4. Can someone explain how (C) relates?
   In particular, why does (C) refer to a functor but (B) refer to a natural transformation? I assume $S \otimes_R S$ is the relevant object in the category of $(S,S)$-bimodules; what is the relevant bit corresponding to the category of functors $\text{Mod}(S) \times \text{Mod}(S) \rightarrow \text{Ab}$? Is the fact that it sends isomorphisms to isomorphisms just rephrasing the statement?

I'd really appreciate some help in understanding this. I'm happy to clarify anything that isn't clear.