I am reading a paper in which the author defines $\mathbf{Mod}$ as the category of pairs $(R, M)$ where $R$ is a ring and $M$ an $R$-module, with morphisms consisting of a ring homomorphism $f$ and an $f$-equivariant module map.
What is meant here by an "$f$-equivariant module map"?
My guess is that if $f: R \to R'$ is a ring homomorphism, it means a linear map $g: M \to M'$ where $M$ is an $R$-module and $M'$ an $R'$-module such that for each $r \in R$, we have $f(r)\cdot g(x) = g(r \cdot x)$ for each $x \in M$.
Is this correct?
That's what it means yes. You can see it as follows : given $f:R\to R'$, and $M$ an $R'$-module, you can define an $R$-module $f^*M$ (this is not a standard notation, as far as I know) as follows : its underlying group is the same as $M$, and the action is the composite $R\times M\to R'\times M\to M$, i.e. $r\cdot m = f(r)m$.
With this, an arrow $(M,R)\to (M',R')$ becomes a ring morphism $f:R\to R'$ together with an $R$-linear map $M\to f^*M'$, which is what they call an $f$-equivariant module map