Given two $R$-modules $M$ and $N$, if there exists an injective module homomorphism $$\phi: M \hookrightarrow N,$$ then we can say $(M, \phi)$ is a submodule of $N$, essentially, $\phi(M)\subset N$. Is this understanding correct?
For example, given the short exact sequence in the first line, we see it is isomorphic to the second exact sequence since the two squares commutes, so here we can view $(\mathbb Z, \times 2)$ as a submodule of $\mathbb{Z}$. $$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z} @>\times 2>> \mathbb{Z} @>\mod 2>> \mathbb{Z}/2 @>>> 0\\ & @VV\times 2V @VV\text{id} V @VV\text{id} V \\ 0 @>>> 2\mathbb{Z} @>\text{inclusion}>> \mathbb{Z} @>\mod 2>> \mathbb{Z}/2 @>>> 0\\ \end{CD}$$
It's a matter of convention what exactly you want "submodule" to mean. More generally, consider what you might want subobject to mean in a category. One definition is that a subobject of an object $X$ is a monomorphism $S \hookrightarrow X$; this is my preferred definition. With this definition the collection of all subobjects of $X$ forms a preorder, which is like a poset but can have objects which are different but isomorphic.
Some people really want the collection of all subobjects to form a poset, and accordingly they define a subobject to be an isomorphism class of monomorphisms $S \hookrightarrow X$; this amounts to taking a skeleton of the preorder above, and has the nice property that if the ambient category is sets, groups, modules, etc. this definition recovers exactly the usual notion of subset, subgroup, submodules, etc. (which form a poset and not a preorder).