Let $R$ be a ring and $I\subset R$ an ideal. Show that $0\to M\to N\xrightarrow{\pi} R/I\to 0$ splits if and only if there exists $x\in N$ with $\pi(x)=1+I$ and $rx=0$ for all $r\in I$.
$\boxed{\Rightarrow}$ There exists a section $s:R/I\to N$ such that $\pi\circ s=\operatorname{id}_{R/I}$. Let $x:=s(1+I)$. We have $rx=s(r+I)=s(I)=0$ for all $r\in I$.
$\boxed{\Leftarrow}$ We want to define a section $s:R/I\to N$ such that $\pi\circ s=\operatorname{id}_{R/I}$. We define $s(1+I)=x$. We extend this defintion by $R$-linearity. This is well defined since $s(r+I)=r\cdot s(1+I)=rx=0$?
Maybe I don't understand the notation $1+I$ for an equivalence class, but what goes wrong here?
Consider $\rho\colon R\to R/I$, the canonical map. Then you can define $\varphi\colon R\to N$ by $\varphi(r)=rx$.
Since $I\subseteq\ker\varphi$ by assumption, the map factors through $\rho$: there exists a unique $s\colon R/I\to N$ such that $s\rho=\varphi$.
Clearly $s(1+I)=s\rho(1)=\varphi(1)=x$.