Let $$0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0$$ be an exact sequence of $R$-modules. Then $M$ is noetherian if and only if $M'$ and $M''$ are.
In my attempt of proving this I didn't use the ascending chain condition, but tried to use the "every submodule of $M$ is finitely generated" condition. I wonder if my proof is correct, since I nowhere used the fact that $f$ is injective?
Assume $M'$ and $M''$ are noetherian. Then $g^{-1}(N)$ is a finitely generated submodule of $M'$ with generators $m_1', \dots m_s' \in g^{-1}(N)$. Let $N\subset M$ be an arbitrary submodule of $M$. Since $g$ is surjective, $g(N)$ is a submodule of $M''$. Since $M''$ is noetherian, we find generators $m_1'' , \dots, m_t'' \in g(N)$ of $g(N)$. There exist $n_1, \dots, n_t \in N$ with $g(n_i) = m_i''$. Now let $n\in N$. Then $$g(n) = \sum_{i=1}^t \lambda_i g(n_i)$$ for some $\lambda_i \in R$. Hence $n - \sum_{i=1}^t \lambda_i n_i \in \ker g = \operatorname{im } f \implies f(m') = n-\sum_{i=1}^n \lambda_in_i$ for some $m' \in M'$. But since $n - \sum_{i=1}^n \lambda_i n_i \in N$, it follows that $m' \in f^{-1}(N)$. Therefore $m' = \sum_{j=1}^s \mu_j m_j' $ for some $\mu_j \in R$, so $$n= \sum_{j=1}^s \mu_j f(m_j') + \sum_{i=1}^t \lambda_i n_i$$ and $N$ is generated by $f(m_j)$, $n_i$.
The assumption that $f$ is injective or that $g$ is surjective aren't necessary, all that matters is that $M'\to M\to M''$ is exact.
Indeed suppose you know the case for when $f$ is injective, and $g$ is surjective, and let $M'\to M\to M''$ be an exact sequence. Then $0\to M'/\ker(f)\to M\to \mathrm{im}(f)\to 0$ is exact, and $M'/\ker(f)$ is noetherian as a quotient of a noetherian module, $\mathrm{im}(f)$ is noetherian as a submodule of a noetherian module; and the rest follows.
This shows that it's not surprising that you didn't use the injectivity of $f$.
Your proof is weird : who is $N$ ? I assume it's the submodule of $M$ under consideration, but then it should be $f^{-1}(N)$ at the beginning, not $g^{-1}(N)$.
Also you don't need $g$ to be surjective for $g(N)$ to be a submodule of $M''$.
Other than that, your proof is fine, and as noted, you didn't use injectivity of $f$, or surjectivity of $g$, but that's not a problem as I pointed out.