This is exercise 0.1 in Rotman's algebraic topology text. I am mostly fine with the proof but I want to be sure I am properly/rigorously translating from sums of elements to the direct product of subgroups (in the second paragraph).
Let $H$ be a subgroup of an abelian group $G$. If there is a homomorphism $r:G \to H$ with $r(x)=x$ for all $x \in H$, then $G=H \oplus \ker r$. (Hint: If $y \in G$, then $y=r(y)+(y-r(y))$.)
Proof: Write $y=r(y)+(y-r(y))$. Clearly $r(y) \in H$ and $r(y-r(y))=r(y)-r(r(y))=r(y)-r(y)=0$, so $y-r(y) \in \ker r$.
Letting $r(y)=(r(y),0)$ and $y-r(y)=(0,y-r(y))$, we have $y=r(y)+(y-r(y))=(r(y),0)+(0,y-r(y))=(r(y),y-r(y))$, and we see that $G=H \oplus \ker r$.
Does this work? Thanks.
The content of the statement $G = H \oplus \mathrm{ker}(r)$ is that each element of $G$ can be written uniquely as a sum of an element from $H$ and an element from $\mathrm{ker}(r)$. You indeed show that $y = r(y) + (y - r(y))$, and that $r(y) \in \mathrm{ker}(h)$.
If you want to be explicit about uniqueness, begin by assuming that $y = a + b$ with $a \in H$, $b \in \mathrm{ker}(r)$. Then $r(y) = r(a) = a$, and thus $b = y - r(y)$, which shows that your way of writing $y$ as a sum of an element from $H$ and one from $\mathrm{ker}(r)$ is also unique.