If we have the following exact sequences of $R-$modules
$0 \rightarrow M_1 \stackrel{f} \rightarrow N \stackrel{g} \rightarrow L_1 \rightarrow 0$
$0 \rightarrow M_2 \stackrel{h} \rightarrow N \stackrel{k} \rightarrow L_2 \rightarrow 0$
and we know that $k \circ f$ is a R-isomorphism
How can we show that $g \circ h$ is also a R-isomorphism?
Well, I just played with your sequences and here is my partial solution.
We need to show two things. First, that $\ker(g\circ h)=\{0\}$ and second, that $g\circ h$ is surjective.
To show that $\ker(g\circ h)=\{0\}$ we chose an element $x\in \ker(g\circ h)$, so $(g\circ h)(x)=g(h(x))=0$, this means that $h(x)\in \ker(g)=\text{Im}(f)$, so there is some $y\in M_1$ such that $h(x)=f(y)$. Applying $k$ to both sides we get $k(h(x))=k(f(y))$, but $h(x)\in \text{Im}(h)=\ker(k)$, thus $k(f(y))=k(h(x))=0$. By hypothesis $k\circ f$ is iso, hence $y=0$ and this gives us that $h(x)=f(y)=f(0)=0$. Since $h$ is injective, we conclude that $x=0$, as desired.
To show that $g\circ h$ is surjective, you have to use your hypothesis and "play" like I did to prove that $\ker(g\circ h)=\{0\}$.