Why an R-module is an R/I-module precisely when it is annihilated by I?

Let $I$ be an ideal of $R$ and let $M$ be an $R$-module. Prove that the formula $(r + I) \cdot m = r \cdot m$ for $r + I \in R/I$, and $m \in M$ makes $M$ into an $R/I$-module if and only if $i \cdot m = 0$ for all $i \in I, m \in M$.

I don't quite understand why we must have $i \cdot m = 0$ instead of $I \cdot m \in M$. This is for proving the only if part.

A $R$-module is nothing else but a ring morphism $R \to End(M)$. By the homomorphism theorem, this induces a ring morphism $R/I \to End(M)$ iff $IM=0$.

$M$ is an $R$-module, so $I\cdot m\in M$ is always true since $I\subseteq R$. So that condition would not really be saying anything useful.

The point here is that if $i\in I$, then $r+I=(r+i)+I$. So for the formula $(r+I)\cdot m=r\cdot m$ to be well-defined, you should have $r\cdot m = (r+i)\cdot m$ for any $i\in I$, and so $r\cdot m = r\cdot m+i\cdot m$ and thus $i\cdot m=0$.