Let $M$ be an $R$-module and $N$ be a submodule. Let $f:M\rightarrow M'$ be a surjective homomorphism with kernel $N$. How to show that $M'$ is isomorphic to $M/N$?
My thought: Define $g:M/N\rightarrow M'$ by $g(a+N)=f(a)$.
Then $$g((a+N)+(b+N))=g(a+b+N)=f(a+b)=f(a)+f(b)=g(a+N)+g(b+N)$$ $$g((a+N)r)=g(ar+N)=f(ar)=f(a)r=g(a+N)r$$
So $g$ is a homomorphism.
For any $a\notin N$, $g(a+N)=f(a)\ne0$. Therefore the kernel of $g$ is $N$ only.
Hence we conclude that $g$ is an isomorphism and hence $M'\approx M/N$.
Is my proof correct?
You defined the map correctly, then you went on to show that it was additive, linear, surjective and injective (all correctly!), but you failed to show that it was well-defined in the first place!
Immediately after you defined the map, you need to show it's well-defined since we're dealing with cosets and their representatives. This amounts to supposing that $a + N = b + N$ and showing that $g(a + N) = g(b + N)$. If this isn't true, then we aren't dealing with a well-defined function (so everything you did afterwards would be meaningless).
Since $a + N = b + N$, then $a - b \in N$ by property of cosets which means that $f(a - b) = 0$ since $N = \ker f$ by hypothesis. Hence $f(a) = f(b)$ and thus we have $$g(a + N) = f(a) = f(b) = g(b + N)$$ as desired.