Let $\mathbb{Z}_k^\times$ be a multiplicative group modulo $k$. Assuming $k=2^c$ is a power of two integer (for positive int $c$), $\mathbb{Z}_k^\times$ consists of odd integers less than $k$.
Does this hold $\mathbb{Z}_k^\times \cong \mathbb{Z}_{2} \times \mathbb{Z}_{\frac{k}{4}}$ and how can we prove that?
Why can't we have the following isomorphism $\mathbb{Z}_k^\times \cong \mathbb{Z}_{2^a} \times \mathbb{Z}_{\frac{k}{2^{a+1}}}$ for some integer $a$?
Consider $c \ge 3$.
Note that $|\mathbb Z_k^\times| = \dfrac k2$.
Since $\mathbb Z_k^\times$ is abelian, we just need to show the following:
For 1, we know that $\mathbb Z_k^\times$ is cyclic, or equivalently, there are primitive roots mod $k$, iff $k = 2$ or $4$ or $p^n$ or $2p^n$. Therefore for $c\ge3$, $\mathbb Z_k^\times$ is not cyclic.
For 2, we need look no further than $3$. The multiplicative order of $3$ mod $k$ is equal to $\dfrac k4$. Here is a proof of this fact: Proof $\operatorname{ord}_{2^n}(3)=2^{n-2}$. This also answers your second question: $a$ can only equal to $1$ since there is an element of order $\dfrac k 4$.
Now the result follows from the Fundamental Theorem of Finite Abelian Groups.