Let $X$ be the compact Riemann surface and $H$ a finite subgroup of $G=Aut(X)$. Then you can think of two different constructions.

1. One is taking the quotient in the category of complex manifolds $X/H$ (which corresponds to the GIT quotient $X//H$ in the algebraic category). Quotients (of complex manifolds by complex Lie groups) don't always exist, so you have to prove that there actually exists a complex manifold $X/H$ which has the desired quotient properties (which is done in your textbook).
2. The other construction is considering the quotient orbifold $[X/H]$. In this case, since $H$ is finite, it has always a meaning, and is -indeed- a 1-dimensional orbifold in the complex manifolds category.

The relation between the two constructions is that (in this case of smooth Riemann surfaces and finite groups) you can think of $[X/H]$ as the Riemann surface $X/H$ decorated, at the ramification points of the quotient map $\pi:X\to X/H$, with the stabilizers.

You say:

"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"

I think the correct statement would be that when $G$ acts freely and properly discontinuously there exists a quotient $M/G$ in the category of manifolds, and the quotient orbifold $[M/G]$ (which is the quotient in the category of orbifolds) happens to be a manifold. But there are cases, like the case of $M=\mathbb{C}\mathbb{P}^1$ and $G=${$1,-1$}, in which the action is not free, nevertheless the quotient manifold $M/G$ does exist (in this case is $\mathbb{C}\mathbb{P}^1$ itself) but is different from the quotient orbifold $[M/G]$ (in this case $[\mathbb{C}\mathbb{P}^1/H]$ is the Riemann sphere decorated at $0$ and $\infty$ with copies of $\mathbb{Z}/2\mathbb{Z}$).

Let $X$ be the compact Riemann surface and $H$ a finite subgroup of $G=Aut(X)$. Then you can think of two different constructions.

1. One is taking the quotient in the category of complex manifolds $X/H$ (which corresponds to the GIT quotient $X//H$ in the algebraic category). Quotients (of complex manifolds by complex Lie groups) don't always exist, so you have to prove that there actually exists a complex manifold $X/H$ which has the desired quotient properties (which is done in your textbook).
2. The other construction is considering the quotient orbifold $[X/H]$. In this case, since $H$ is finite, it has always a meaning, and is -indeed- a 1-dimensional complex-analytic orbifold.

The relation between the two constructions is that (in this case of smooth Riemann surfaces and finite groups) you can think of $[X/H]$ as the Riemann surface $X/H$ decorated, at the ramification points of the quotient map $\pi:X\to X/H$, with the stabilizers.

You say:

"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"

I think the correct statement would be that when $G$ acts freely and properly discontinuously there exists a quotient $M/G$ in the category of manifolds, and the quotient orbifold $[M/G]$ (which is the quotient in the category of orbifolds) happens to be a manifold. But there are cases, like the case of $M=\mathbb{C}\mathbb{P}^1$ and $G=${$1,-1$}, in which the action is not free, nevertheless the quotient manifold $M/G$ does exist (in this case it's $\mathbb{C}\mathbb{P}^1$ itself) but is different from the quotient orbifold $[M/G]$ (in this case $[\mathbb{C}\mathbb{P}^1/H]$ is the Riemann sphere decorated at $0$ and $\infty$ with copies of $\mathbb{Z}/2\mathbb{Z}$).

Let $X$ be the compact Riemann surface and $H$ a finite subgroup of $G=Aut(X)$. Then you can think of two different constructions.

1. One is taking the quotient in the category of complex manifolds $X/H$ (which corresponds to the GIT quotient $X//H$ in the algebraic category). Quotients (of complex manifolds by complex Lie groups) don't always exist, so you have to prove that there actually exists a complex manifold $X/H$ which has the desired quotient properties (which is done in your textbook).
2. The other construction is considering the quotient orbifold $[X/H]$. In this case, since $H$ is finite, it has always a meaning, and is -indeed- a 1-dimensional orbifold in the complex manifolds category.

The relation between the two constructions is that (in this case of smooth Riemann surfaces and finite groups) you can think of $[X/H]$ as the Riemann surface $X/H$ decorated, at the ramification points of the quotient map $\pi:X\to X/H$, with the stabilizers.

Let $X$ be the compact Riemann surface and $H$ a finite subgroup of $G=Aut(X)$. Then you can think of two different constructions.

1. One is taking the quotient in the category of complex manifolds $X/H$ (which corresponds to the GIT quotient $X//H$ in the algebraic category). Quotients (of complex manifolds by complex Lie groups) don't always exist, so you have to prove that there actually exists a complex manifold $X/H$ which has the desired quotient properties (which is done in your textbook).
2. The other construction is considering the quotient orbifold $[X/H]$. In this case, since $H$ is finite, it has always a meaning, and is -indeed- a 1-dimensional orbifold in the complex manifolds category.

The relation between the two constructions is that (in this case of smooth Riemann surfaces and finite groups) you can think of $[X/H]$ as the Riemann surface $X/H$ decorated, at the ramification points of the quotient map $\pi:X\to X/H$, with the stabilizers.

You say:

"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"

I think the correct statement would be that when $G$ acts freely and properly discontinuously there exists a quotient $M/G$ in the category of manifolds, and the quotient orbifold $[M/G]$ (which is the quotient in the category of orbifolds) happens to be a manifold. But there are cases, like the case of $M=\mathbb{C}\mathbb{P}^1$ and $G=${$1,-1$}, in which the action is not free, nevertheless the quotient manifold $M/G$ does exist (in this case is $\mathbb{C}\mathbb{P}^1$ itself) but is different from the quotient orbifold $[M/G]$ (in this case $[\mathbb{C}\mathbb{P}^1/H]$ is the Riemann sphere decorated at $0$ and $\infty$ with copies of $\mathbb{Z}/2\mathbb{Z}$).