I have commutative rings $A$, $B$ and $C$, and unital ring homomorphisms $f:A\to B$, $g:A\to C$. Using $f$ and $g$, we can regard $B$ and $C$ as $A$-modules, and thus define the $A$-module $B\otimes_A C$, which can also regarded as a $B$-module or a $C$-module.
I have read that such a construction gives an endofunctor of $C$-modules along the lines of $B\otimes_{A,g}(-)$ but I am struggling to see precisely how this works (what might $\otimes_{A,g}$ even mean?)
Start with a $C$-module $M$. Using $g$, we can regard $M$ as an $A$-module. Then $B$ is an $A$-module via $f$, and $B\otimes_A M$ is an $A$-module. But how do I obtain a $C$-module from this?
I would guess that for a $C$-module $M$ the term $B \otimes_{A,g} M$ is a short hand for $(B \otimes_A C) \otimes_A M_A$, where $M_A$ denotes the $A$-module $M$ (restriction of the module structure along $g$) and the $C$-module structure is given by multiplication in the middle $C$.
In other words, you should obtain an endo-functor $$\mathsf{Mod}_C \overset{U}\rightarrow \mathsf{Mod}_A \overset{(B\otimes_A C)\otimes_A -}\longrightarrow \mathsf{Mod}_{B\otimes_A C} \overset{U}\rightarrow \mathsf{Mod}_C$$ where the $U$'s are forgetful functors and the functor in the middle is change of base / extension of scalars.
This is a blind guess though, without more context it is hard to tell what your notation explicitly means.