First and foremost, some definitions. Every modules will be assumed right $R$ modules. We say that a submodule $N$ of $M$ is coneat in $M$ if for each simple module $S$, every module morphism $N \to S$ can be extended to a morphism $M \to S$. We say that a submodule $N$ of $M$ is coclosed in $M$ if $N/K \ll M/K$ implies that $K=N$ for each submodule $K$ of $N$. Recall that $\ll$ means superfluous submodule.
Last but not least, $M$ is called coatomic if every proper submodule $N$ of $M$ is contained in a maximal submodule of $M$. What I'm aiming to prove is the following claim:
Let $M$ be a module and $N$ be a coatomic module of $M$. Then $N$ is coneat in $M$ if and only if $N$ is coclosed in $M$.
For the first implication let's suppose that $N$ is coneat in $M$, so I want to prove that $N$ is coclosed by contradiction. So let's suppose that $N$ is coneat but it is not coclosed, this is, $N/X \ll M/X$ for some proper submodule $X \leq N$ ($X \neq N$). Since $N$ is coatomic, then $X$ is contained in a maximal submodule, let's say $K$ of $N$. But I don't know how to make further progress with the proof. For the other implication I'm run out of ideas. I mean I need to suppose there is a morphism $N \to S$ where $S$ simple but I need to extend this morphism to another $M \to S$. Somehow I know that if $S$ is simple then its the quotient of a maximal module but I don't know how to apply our hypothesis.
For the first implication, we can continue as follows. We are assuming that $N$ is coneat, that $N/X$ is superfluous in $M/X$, and that $X$ is contained in a maximal submodule $K$ of $N$. Now $S=N/K$ is simple, and by assumption the canonical map $N\to S$ lifts to a map $M\to S$, say with kernel $U$. Then $U$ is a maximal submodule of $M$, and $U\cap N=K$. Thus $U+N=M$, so $U/X+N/X=M/X$, which by assumption implies that $U/X=M/X$, and hence that $U=M$, a contradiction. Thus $N$ is coclosed in $M$.
For the other implication, suppose that $N$ is coclosed but not coneat, so there exists a simple module $S$ and a map $N\to S$, say with kernel $K$, which does not extend to $M$. Now $N/K$ is not superfluous in $M/K$, so there exists a proper submodule $U$ of $M$ containing $K$ such that $U+N=M$. Then $U\cap N$ is not equal to $N$, so equals $K$, and hence $M/U\cong N/N\cap U=N/K=S$ is simple, so $U$ is maximal, and the map $M\to M/U=S$ extends $N\to N/K=S$, a contradiction. Thus $N$ is coclosed.