So i have been reading a little bit about riemann surfaces and at some point the author says that if a group $G$ acts on $X$, our riemann surface, by holomorphic automorphisms, such that :
1) Around each point $p \in X$ we can find an open neighborhood $N$ such that if $q_1,q_2 \in N$ and $g.q_1=q_2$ then $g=id$ and $q_1=q_2$.
2) Suppose $p,q \in X$ wich do not lie in the same orbit, then there are neighborhoods $N_1,N_2$ of $q_1,q_2$ such that $GN_1$ is disjoint from $N_2$.
Now i am trying to see that in fact $X/G$ is a Riemann surface. I think condition 2) will give us the fact that the space is hausdorff, and 1) says that for every element we can find a neighborhood where the projection is a bijection and these are going to be the open sets to construct our riemann surface, but i cant seem to find the coordinate charts, i know i can find coordinate charts in $X$ and i can send $N$ to $X$ but this open set can be partitioned into the different open sets that give $X$ the structure of a riemann surface. So any tips would be appreciated , thanks.