I need to do the following without just applying Yoneda Lemma, and I am struggling to see what to do.
Let $f: A \to B$ and $g: B \to C$ be arrows in an abelian category $\mathcal{A}$. If $Hom(X,A) \xrightarrow{Hom(f)} Hom(X,B)$ is the kernel of $Hom(X,B) \xrightarrow{Hom(g)} Hom(X,C)$ in the category of abelian groups for any $X \in \mathcal{A}$, then $f: A \to B$ is the kernel of $g: B \to C$
Use the universality property of a kernel.
If $$\mathrm{Hom}(X,A)\xrightarrow{\mathrm{Hom}(f)}\mathrm{Hom}(X,B)$$ is the kernel of $$\mathrm{Hom}(X,B)\xrightarrow{\mathrm{Hom}(g)}\mathrm{Hom}(X,C),$$ then $\mathrm{Hom}(g)\mathrm{Hom}(f)=\mathrm{Hom}(gf)$ is the zero map and for any map $h:H\to\mathrm{Hom}(X,B)$ such that $\mathrm{Hom}(g)h=0$, there exists a unique map $u:H\to\mathrm{Hom}(X,A)$ such that $\mathrm{Hom}(f)u=h$.
Take $X=A$ and see that $0=\mathrm{Hom}(gf)(\mathrm{id}_{A})=gf$, and for any map $h:H\to B$ such that $\mathrm{Hom}(g)h=gh=0$, there exists $u:H\to A$ such that $\mathrm{Hom}(f)u=fu=h$. Thus the universal property is satisfied for $$A\xrightarrow{f}B\xrightarrow{g}C.$$