For integers $h$ odd and $k$ arbitrary I have $n = h\cdot 2^k$. Let's assume a primepower $p^{\alpha}$ divides $n$, but not $h\cdot 2^{k-1}$. Then
$$p^{\alpha}\mid h\cdot 2^k \Leftrightarrow p^{\alpha}\mid h \text{ or } p^{\alpha}\mid2^{k}$$ and $$ p^{\alpha}\nmid h\cdot 2^{k-1} \Leftrightarrow p^{\alpha}\nmid h \text{ and } p^{\alpha}\nmid 2^{k-1}.$$
With propositional logic we get $$ p^{\alpha}\mid h\cdot 2^k \text{ and } p^{\alpha}\nmid h\cdot 2^{k-1} \Leftrightarrow (p^{\alpha}\mid h \text{ and } p^{\alpha}\nmid h \text{ and } p^{\alpha}\nmid 2^k) \text{ or } (p^{\alpha}\mid 2^k \text{ and } p^{\alpha}\nmid h \text{ and } p^{\alpha} \nmid 2^{k-1} ) \Leftrightarrow (p^{\alpha}\mid 2^k \text{ and } p^{\alpha}\nmid h \text{ and } p^{\alpha} \nmid 2^{k-1} ). $$ I can conclude that $p^{\alpha} = 2^k.$
But what can I conclude for a composite number $a$ such that $$a\mid h\cdot 2^k \text{ and } a\nmid h\cdot 2^{k-1}?$$
I tried this with $a = p\cdot q$, but even if it worked, what it to be done for $a = \prod\limits_p p^{\alpha_p}$, where this is the prime decomposition of $a$?
You implicitely assumed $k\ge1.$
If $a\mid2^kh$ and $a\nmid2^{k-1}h$ then $a$ is even, $a=2b,$ $b\mid2^{k-1}h$ and, in case $k\ge2,$ $b\nmid2^{k-2}h.$ Iterating, we eventually get: $$a=2^kc,\quad c\mid h.$$
Conversely, every integer $a$ of this form satisfies the two conditions.