A group is generated by two elements of order $2$ is infinite and non-abelian

(a) The map $\phi$ is a homomorphism by property of the free group (it is actually its universal property, anytime you give the image of the generators you actually define a homomorphism).

(b) Actually the definition of a group by generators and relations (such as the one for $G$ here) is exactly that $G:=\mathbb{F}_{\text{generators}}/<<\text{relations}>>$.

In this setting, the homomorphism $\phi$ is actually defined as the canonical projection of $\mathbb{F}_{\text{generators}}$ on its quotient $G$. Hence $Ker(\phi)=<<\text{relations}>>=H$ by definition (or more precisely by the universal property of the quotient).

(c) To me, you have the good direction. Now what you need to show is that $(xy)^k\in H$ iff $k=0$. What I would show (and I think this easy) is the following assertion :

Any reduced form of any non-trivial word of $H$ contains at least one occurence of $x^{2l}$ or $y^{2l}$ for some $l\neq 0$.

Since $(xy)^k$ is never trivial for $k\neq 0$ and the absolute values of the powers of $x$ and $y$ in a reduced form of $(xy)^k$ are $1$ then we know that $(xy)^k\notin H$.