I am trying to solve the following problem.
Let $M$ be a module over a commutative ring $A$ with identity. $M'\subseteq M$ is a simple submodule. Also, $N=M/M'$ is simple. Assume that $N$ is not isomorphic to $M'$.
Prove that $M\cong M'\oplus N$
Hint: Look at $m_0\in M$ such that $M'+m_0$ generates $N$ and consider $\langle m_0\rangle$.
This is what I have done so far:
Since $N=\langle \overline{m_0}\rangle$ for all $m\in M$ we have $$\overline{m}=c_m\overline{m_0}$$ for some $c_m\in A$.
This means that $m-c_mm_0\in M'$ for all $m$.
Since $M'$ is simple we have $M'=\langle m'\rangle$ for some $m'\in M$.
Since $m-c_mm_0\in M'$ it follows that $$m-c_mm_0=d_mm'$$ for some $d_m\in A$. So $$m=c_mm_0+d_mm'$$
It follows that $$M=M'+\langle m_0\rangle$$
From here I don't know what to do next. Any ideas?
Let us write $N'=\langle m_0\rangle$. You have shown $M'+N'=M$. If you knew that $M'\cap N'=0$, then $M$ would be the direct sum of the submodules $M'$ and $N'$, and you would be done since then $N'\cong N$ via the projection map (I'll leave verifying the details to you).
Unfortunately, $M'\cap N'$ may not be trivial. If it's not trivial, though, it must be all of $M'$, since $M'$ is simple. In that case, then, $M'\subseteq N'$ so $M=M'+N'=N'$. In particular, since $N'$ is cyclic, this means $M$ is cyclic.
Since $M$ is cyclic, it is isomorphic to $A/K$ where $K$ is the annihilator of $M$. So let's try to identify this ideal $K$. Let $I$ be the annihilator of $M'$ and $J$ be the annihilator of $N$. Since these modules are simple and not isomorphic, $I$ and $J$ are maximal ideals and $I\neq J$, so in particular $I+J=A$.
Clearly $K\subseteq I\cap J$ (the annihilator of a submodule or quotient can only be larger). On the other hand $IJ\subseteq K$, since if $i\in I$ and $j\in J$, then $jm_0\in M'$ (since $j$ annihilates $N$) and then $ijm_0=0$ (since $i$ annihilates $M'$).
So $IJ\subseteq K\subseteq I\cap J$. But since $I+J=A$, by the Chinese remainder theorem $I\cap J=IJ$ and therefore $K=I\cap J$. Also by the Chinese remainder theorem, $A/(I\cap J)\cong A/I\oplus A/J$. That is, $M\cong M'\oplus N$.