Images of composed homomorphisms Let $R$ be a ring and $M,N,S$ be $R$-modules. Let $\varphi_1 : M\to N$ and $\varphi_2 : N\to S$ be homomorphisms.
Then $\operatorname{im}(\varphi_2\circ\varphi_1) \subseteq \operatorname{im}(\varphi_2)$ and $\ker(\varphi_1) \subseteq \ker(\varphi_2\circ\varphi_1)$, so we may form the quotients $\operatorname{im}(\varphi_2)/\operatorname{im}(\varphi_2\circ\varphi_1)$ and $\ker(\varphi_2\circ\varphi_1)/\ker(\varphi_1)$.
For the kernels, the application of the first isomorphism theorem to the homomorphism $\varphi_1|_{\ker(\varphi_2\circ \varphi_1)} : \ker(\varphi_2\circ \varphi_1) \to N$ yields the nice description
$$
\ker(\varphi_2\circ\varphi_1)/\ker(\varphi_1) \cong \operatorname{im}(\varphi_1) \cap \ker(\varphi_2).
$$
Now my question is:

Is there a similar description for the quotient
  $$\operatorname{im}(\varphi_2)/\operatorname{im}(\varphi_2\circ\varphi_1)?$$

I didn't succeed in finding a suitable application of the first isomorphism theorem in this case.
We may observe that the above reasoning for the kernels works exactly the same for homomorphisms of groups. However for the quotient of the images to make sense for groups, we need the additional property $\operatorname{im}(\varphi_2\circ\varphi_1)$ is a normal subgroup of $\operatorname{im}(\varphi_2)$.
This indicates that the situation might be a bit different for the quotient of the images. If needed, please assume some additional property on the ring $R$ to get a meaningful answer.

For ease of notation, let $A = \operatorname{im}\varphi_1 \subset N$, $T = \operatorname{im}\varphi_2 \subset S$, and $Q = T/\operatorname{im} (\varphi_2\circ\varphi_1) = T/\varphi_2(A)$. Consider the map $\psi = \pi \circ \varphi_2 \colon N \to Q$, where $\pi \colon T \to Q$ is the canonical projection.
The kernel of $\psi$ is
$$\ker \psi = \psi^{-1}(0) = \varphi_2^{-1}(\pi^{-1}(0)) = \varphi_2^{-1}(\varphi_2(A)) = A + \ker\varphi_2,$$
and $\psi$ is surjective because $\varphi_2$ and $\pi$ are ($\varphi_2$ considered as a homomorphism to $T$). By the homomorphism theorem, we have an induced isomorphism
$$\overline{\psi} \colon \frac{N}{\operatorname{im}\varphi_1 + \ker \varphi_2} \to Q = \frac{\operatorname{im}\varphi_2}{\operatorname{im}(\varphi_2\circ\varphi_1)}.$$
