Given a ring homomorphism $f:S \to R$ one can see that every right $R$-module $M$ can be seen as a right $S$-module. Then one can ask himself if given a module homomorphism $\phi:M_R \to N_R$ this induces a module homomorphism $\phi:M_S \to N_S$.
I suspect that what my professor is looking here is for a functor. What my notes say is that imposing the compatibility relation $ms = mf(s)$ I should get that $\phi(ms) = \phi(mf(s)) = \phi(m)f(s) = \phi(m)s$.
However, the last step is unclear to me. Is it not too strong to impose the condition $ms = mf(s)$?
Thanks to @Max answer I realize what is happening here.
Since $M$ has a module structure on $R$, there is a ring homomorphism $R \to End(M)$. Using $f$ we get a ring homomorphism $S \to R \to End(M)$. But using this ring homomorphism implies defining $ms = mf(s)$ which also solves the two questions asked.