Suppose we have $A,B$ which are $R$-submodules of the $R$-module M. Then , there exists an exact sequence :
$0\rightarrow \frac{M}{A\cap B}\rightarrow \frac{M}{A}\times \frac{M}{B}\rightarrow \frac{M}{A+B} \rightarrow 0$
Where $A+B=\langle A\cup B\rangle$
I believe that that the first function should be something like $α(u)=(u,u)$ and the second one $β(u,v)=u-v$ to satisfy that ${\rm im}\;\alpha=\ker\beta$ but I can't show surjectivity of $β$. Any help will be appreciated.
The first map must take a class in $M/(A\cap B)$ and give a pair of classes in $(M/A)\times(M/B)$. Clearly $$([x]_{A \cap B})\mapsto \left([x]_A,[x]_B \right)$$will do. One must check that this is injective. If $[x]_A = [0]_A$ then $x \in A$. Similarly $[x]_B=[0]_B$ implies $x \in B$. So $x \in A \cap B$ means $[x]_{A \cap B} = [0]_{A \cap B}$.
Procceding, we must have a map $(M/A)\times (M/B) \to M/(A+B)$. Naturally we'll try $$([x]_A,[y]_B)\mapsto [x-y]_{A+B}.$$This is well-defined, since if $[x]_A = [x']_A$ and $[y]_B = [y']_B$, then $x-x' \in A$ and $y-y' \in B$, hence $(x-y)-(x'-y') \in A+B$. We only have to check that this map is surjective and its kernel of this map is the image of the previous one.
For surjectiveness, if $[z]_{A+B}$ is given, then $([z]_A,[0]_B)$ is sent to $[z]_{A+B}$. Finally, $([x]_A,[x]_B)$ is clearly sent to $[0]_{A \cap B}$.
And if $([x]_A,[y]_B)$ is sent to $[0]_{A +B}$, we must check that they can be both represented by a class in $M/(A \cap B)$. We have $x-y \in A+B$, meaning that $x - y = a+b$ for $a \in A$ and $b \in B$, so that $x-a = y+b \in A \cap B$. Meaning that $$[x-a]_A = [y+b]_A = [x]_A \quad \mbox{and}\quad [x-a]_B = [y+b]_B = [y]_B,$$ as wanted.