Suppose $G$ is a group such that Sylow-$2$-subgroup is cyclic. Then can it surject into $\mathbb{Z}_2\oplus\mathbb{Z}_2$?
If $G$ is an abelian group and Sylow-2-subgroup is Cyclic, then it can not, by the Structure theorem of Abelian groups.
Is it true in general? Is it true in the case of $H\rtimes\mathbb{Z}_{2^k}$, when $|H|$ is odd.
Note that in case the Sylow-$2$-subgroup is cyclic and $|G|=2^km$ for some odd $m$, then $G$ has a normal subgroup of order $m$, say $H$. Since $(|H|,2^k)=1$, this implies that $G=H\rtimes\mathbb{Z}_{2^k}$.
Certainly it is impossible in the case of $H\rtimes\mathbb{Z}_{2^k}$ for $|H|$ odd: if we restrict a homomorphism to $f:H \rtimes \Bbb Z_{2^k} \to \Bbb Z_2 \oplus \Bbb Z_2$ to $H$, it will be trivial, hence $f$ is surjective iff its restriction to $\Bbb Z_{2^k}$ is surjective.
As you noted, an application of Schur-Zassenhaus shows that all possible finite groups with cyclic $2$-Sylow are of this form, so this answers the question.