In Algebra: Chapter $0$ by Aluffi, at the end of page 161, he writes:
Since every submodule $N$ is then the kernel of the canonical projection $M \to M/N$, our recurring slogan becomes, in the context of $R$-$\mathsf{Mod}$ $$\text{kernel} \iff \text{submodule}:$$ unlike as in $\mathsf{Grp}$ or $\mathsf{Ring}$, being a kernel poses no restriction on the relevant substructures. Put otherwise, ‘every monomorphism in $R$-$\mathsf{Mod}$ is a kernel’; this is one of the distinguishing features of an abelian category.
I know what a monomorphism is:
In a category $\mathsf{C}$, a morphism $f: A \to B$ is a monomorphism if for any object $Z$ and any two morphisms $\alpha: Z \to A$, $\alpha': Z \to A$, we have $f\alpha = f\alpha' \implies \alpha = \alpha'$.
What does 'every monomorphism in $R$-$\mathsf{Mod}$ is a kernel' mean?
Suppose we have a morphism $\phi:X\to Y$ with kernel $K$. In Aluffi, the emphasis is placed on the domain $X$, but the kernel is really both the object $K$ as well as the (mono)morphism $K \hookrightarrow X$.
Remember that in $\mathbf{Grp}$, $\mathbf{Ring}$, and $R–\mathbf{Mod}$, monomorphism $\iff$ injective. So an equivalent formulation to the statement in your question is that every injective morphism's domain is the kernel of some homomorphism.
The reason this is true in $R–\mathbf{Mod}$ is that given the morphism $X \to X/K$, its kernel is $K$! While in group, this is not necessarily the case, because $X/K$ is not a group if $K$ isn't normal. Thus, in $R–\mathbf{Mod}$ (and in $\mathbf{Ab}$), any injection's domain is the kernel of some morphism, but this is not true in general in $\mathbf{Grp}$.