Let $X$ be a hyperelliptic Riemann surface, and let $J$ be the hyperelliptic involution. Then consider the quotient surface $X/ < J > ,$ my question is whether $X/ < J > $ is a Riemann surface?

On the standard Riemann surface textbook, the answer is yes, $X/< J >$ is the Riemann sphere $S^{2}$. More generally, let $H$ be a subgroup of the automorphism group of a Riemann surface $\Sigma,$ then $\Sigma/H$ is also a Riemann surface, and the quotient map is holomorphic, and the fixed points of $H$ are the branch points of the quotient map.

However, when we consider the problem in another way, the above mentioned $X/< J >$ should be an orbifold. The fixed points of $J$ are the singular points of orbifold. An orbifold is not a manifold, and if $X/< J >$ is even not a manifold, how could it be a Riemann surface? We know that only when an group $G$ act freely and properly discontinuously on a manifold $M$, the quotient space $M/G$ is a manifold. Here the question is that the action of automorphism group of a Riemann surface on itself has the fixed points.

A simple example is the American football model of an orbifold. Let $\tau$ be the $\pi$ rotation around $z$-axis, then we can get an orbifold with two singular points located at north pole and south pole of the sphere. However, under the Riemann surface view, $S^{2}$ is a Riemann surface, and $\tau$ is a holomorphic involution with two fixed points, so $S^{2}/<\tau>$ should be a Riemann surface. It seems we get a contradiction here. Where is the mistake or confusion in my above statement?