= Solution
Let $c=\bigcup G$ and $d(n)=c(2n)$. Define
$$
H=\{p\in\operatorname{Fn}(\omega,2)^M:p\subseteq d\}.
$$
This is a filter: restrictions of finite pieces of $d$ remain in $H$, and the union of two members is a common stronger condition.
To prove genericity, take a dense set $D\in M$. Let $E_D$ consist of conditions $q\in\operatorname{Fn}(\omega,2)$ for which some $p\in D$ satisfies
$$
p(n)=q(2n)\qquad(n\in\operatorname{dom}p).
$$
The set $E_D$ is dense. Indeed, given $q$, first read its finitely many assigned even coordinates as a condition $p_0$ on $\omega$. Choose $p\le p_0$ in $D$, and extend $q$ by setting $q(2n)=p(n)$ at the remaining coordinates of $p$. Since $E_D\in M$, the <generic filter> $G$ meets it. For $q\in G\cap E_D$, the corresponding $p\in D$ is a finite subfunction of $d$, so $p\in H\cap D$.
Thus $H$ meets every dense subset belonging to $M$. Moreover, each singleton $\{(n,d(n))\}$ belongs to $H$, and hence
$$
\boxed{H\text{ is }\operatorname{Fn}(\omega,2)\text{-generic over }M,qquad\bigcup H=d.}
$$
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