Problem: Let $R$ be a comutative ring and $M$ be an $R$-module. Let $N_1$ and $N_2$ be submodules of $M$. If $M/N_1, M/N_2$ are Noetherian then $M/(N_1∩N_2)$ is noetherian.
Solution: Let's take $M/N_1, M/N_2$ be Noetherian and $N:=N_1+N_2$.
Let $S_1⊆S_2⊆⋯⊆M/N$ be a chain of submodules. Let us define
$A_i:=\{m+N_1:m+N∈S_i\}$ for each $i$. Then $A_i$ is a submodule of $M/N_1$. If $m+N_1∈A_i$ then $m+N ∈ S_i ⊆ S_{i+1} \implies m+N_1 ∈ A_{i+1}$.
Now, $A_1 ⊆ A_2 ⊆...⊆ M/N_1$ is a chain of submodules. Next, $M/N_1$ is Noetherian, so for each $a$ with $a ≤ i ≤ k $, we have $A_i = A_k.$ So for each $a$ with $a≤i≤k$ we have $$S_i = \{ m+N: m+N_1 ∈ A_i \} = \{ m+N : m+N_1 ∈ A_k \} = S_k.$$
Now $A_j:=\{ m+N_2 : m+N ∈ S_j \}$ for each $j$, and $A_j$ is a submodule of $M/N_2$. If $m+N_2 ∈ A_j$ then $m+N ∈ S_j ⊆ S_{j+1} \implies m+N_2 ∈ A_{j+1}$.
$A_1 ⊆ A_2 ⊆...⊆ M/N_2$ is a chain of submodules. $M/N_2$ is Noetherian so for each $a$ with $a ≤ j ≤ h$ we have $A_j = A_h$. Hence, for each $a$ with $a ≤j ≤ h$ we get $$S_j = \{ m+N : m+N_2 ∈ A_j \} = \{ m+N : m+N_2 ∈ A_h \} = S_h.$$
Hence $M/N := M/(N_1+N_2)$ is Noetherian. $N_1 ∩ N_2 ⊆ N_1+N_2$, so $M/(N_1∩N_2)$ is Noetherian.
Is everything right in my solution? Please help me.
I suggest another way to approach this problem, because I don't like too much index juggling. Note also that we don't need that $R$ is a commutative ring.
Define the $R$-linear map $$\psi: M\to M/N_1 \oplus M/N_2: m\mapsto (m + N_1, m + N_2).$$ Then $\ker \psi = N_1 \cap N_2$ and by the first isomorphism theorem $$M/(N_1 \cap N_2) \cong \operatorname{Im}(\psi). \quad (*)$$
By assumption, $M/N_1$ and $M/N_2$ are Noetherian, so their direct sum $M/N_1 \oplus M/N_2$ is also Noetherian. Hence, $\operatorname{Im}(\psi)$ is Noetherian, as a submodule of a Noetherian module. In view of the isomorphism $(*)$, it follows that $M/(N_1 \cap N_2)$ is Noetherian.