Consider an automorphism of a hyperelliptic curve. It is sufficient to ask what the quotient by this automorphism can be. Because there is a unique hyperelliptic projection to $\mathbb P^1$, it is canonical, hence the automorphism factors through to an automorphism of $\mathbb P^1$. As a finite order automorphism of $\mathbb P^1$ it fixes two points, and if those points are moved to $0$ and $\infty$ it acts by multiplication by $\mu_n$.

Either the cyclic subgroup generated by this automorphism contains the hyperelliptic involution, or it does not. If it does, then clearly we are in your case. There are only two ramification points so the cover must have the form $y^{n/2}=\frac{(x-\alpha_1)}{(x-\alpha_2)} \prod (x-\beta_i)^{n/2}$, so the full cover has the form $y^n=\frac{(x-\alpha_1)}{(x-\alpha_2)} \prod (x-\beta_i)^{n/2}$.

In the other case, the quotient curve is hyperelliptic or elliptic, and you are taking the fiber product of the hyperelliptic cover with another cyclic cover of $\mathbb P^1$. Thus the cover must have the form $y^n=\frac{x-\alpha_1}{x-\alpha_2}$, where the meromorphic function $x$ is the hyperelliptic projection. Special attention should be devoted to the the case when $\alpha_1$ and $\alpha_2$ are ramification points of the hyperelliptic projection. This will create singularities that must be normalized to get the actual ramified cover. In particular, this can make the cover not actually ramified at those points if $n$ is even. Thus the total number of ramification points can be anything from $0$ to $4$.