Alternatively, you could prove directly that $-4H = \frac{1}{2}H$, using the observation that $-4(-2)^n = (-1)^{n+1}2^{n+2} = (-1)^{n+3}2^{n+2} = \frac{1}{2}(-2)^{n+3}$.
(Alternatively, we can take inverses of both sides of the equation $aH = bH$, where by the inverse of a set $S$ we mean $\{x^{-1} : x \in S\}$, and then use the fact that $H^{-1} = H$.)
The proof that $f$ is injective is similar. Let $a,b\in G$ and assume that $Ha^{-1} = Hb^{-1}$; we want to show that $aH = bH$. Equivalently, assume that $a^{-1} \in Hb^{-1}$; we want to show that $a \in bH$. Take $h \in H$ such that $a^{-1} = hb^{-1}$. Then $a = (a^{-1})^{-1} = (hb^{-1})^{-1} = (b^{-1})^{-1}h^{-1} = bh^{-1} \in bH$, because $h^{-1} \in H$.
$f$ is surjective because for every $a \in G$ we have $f(a^{-1}H) = H(a^{-1})^{-1} = Ha$.
(It is $1$ if $a$ and $b$ are both zero; otherwise it is $p$.)
Alternatively, letting $(a,b) \in \mathbb{Z} \times \mathbb{Z}_2$, we can show directly that $\langle (a,b) \rangle \ne \mathbb{Z} \times \mathbb{Z}_2$: in the case $a = 0$ we have $(1,0) \notin \langle (a,b) \rangle$, and in the case $a \ne 0$ we have $(0,1) \notin \langle (a,b) \rangle$. (For the second case $a \ne 0$, note that if $n(a,b) = (0,1)$ then $na = 0$, so $n = 0$, so $nb = 0$, so $0 = 1$, a contradiction.)