Let $R$ be a PID. Consider a ring homomorphism $R \to S$ that makes $S$ into a free $R$-algebra of rank $n$. Let $M$ be a finite $S$-module and denote by $M_{\mid R}$ its restriction as an $R$-module which is also free of rank $n$. Denote by $M^\#$ the $S$-module $\operatorname{Hom}_R(M_{\mid R},R)$ by multiplication into the argument. The latter means that for $f \in M^\#$ and $s \in S$ we define $(s \cdot f)(m) := f(sm)$.
I have the following question/task:
Assume that there is non-zero $m \in M$ such that for all non-zero $s \in S$ we have $s m \neq 0$. Show that there is an element $f \in M^\#$ such that for all non-zero $s \in S$ we have $s \cdot f \neq 0$.
A sufficient condition: If the intersection of the free $R$-module $mS$ of rank $n$ with $\ker(f)$ (which should be of rank $n-1$ as long as $f \neq 0$) is trivial, we have found $f \in M^\#$ with the intended property. But is it possible that $mS \cap \ker(f) \neq \{0\}$ for all $f \in M^\#$?