Riemann surfaces are one dimensional complex manifolds. It is in fact true that any complex $n$-manifold is orientable as a real manifold. One possible way of showing this, is to calculate the determinant of the jacobians of the transition functions.
Suppose you have an $n$-dimensional complex manifold $M$. That is, $M$ is a (first countable, Hausdorff) space such that every point $p\in M$ has an open neighborhood $V$ homeomorphic to an open subset $\Omega\subset\mathbb C^n$,
$$\phi:V\rightarrow\Omega~\mathrm{homeomorphism}$$
and when two such neighborhoods intersect, the transition functions are holomorphic,
$$\psi\circ\phi^{-1}:\phi(V\cap V')\rightarrow\psi(V\cap V')~\mathrm{biholomorphism.}$$
By identifying $\mathbb C^n$ with $\mathbb R^n$ like so: $(z^1,\dots,z^n)\leftrightarrow(x^1,y^1,\dots,x^n,y^n)$ where $z^k=x^k+iy^k$ is the real part/imaginary part decomposition, we get a real manifold structure on $M$. We can calculate the jacobian matrices of the transition functions for both structures.
Let's set up some notation. The first coordinate chart will be $\phi:V\rightarrow\Omega, p\mapsto (z^1(p),\dots,z^n(p))$, and the second will be $\psi:V'\rightarrow\Omega', p\mapsto (Z^1(p),\dots,Z^n(p)).$ In the complex case, we have
$$\mathrm{Jac}_{\phi(p)}(\psi\circ\phi^{-1})=\left(\frac{\partial~[\psi\circ\phi^{-1}]^k}{\partial~z^l}\right)_{1\leq k,l\leq n}=\left(\frac{\partial~Z^k(z^1,\dots,z^n)}{\partial~z^l}\right)_{1\leq k,l\leq n}\\=\left(c_l^k\right)_{1\leq k,l\leq n}\in\mathrm{GL}(n,\mathbb C),$$
and in the real case you'll find
$$\mathrm{Jac}_{\phi(p)}(\psi^{\mathbb R}\circ(\phi^{\mathbb R})^{-1})=
\left(\begin{array}{cc}
\frac{\partial~X^k(x^1,y^1,\dots,x^n,y^n)}{\partial~x^l} & \frac{\partial~X^k(x^1,y^1,\dots,x^n,y^n)}{\partial~y^l} \\
\frac{\partial~Y^k(x^1,y^1,\dots,x^n,y^n)}{\partial~x^l} & \frac{\partial~Y^k(x^1,y^1,\dots,x^n,y^n)}{\partial~y^l}
\end{array}\right)_{1\leq k,l\leq n}=
\left(\begin{array}{cc}
\mathrm{Re}(c_l^k) & -\mathrm{Im}(c_l^k) \\
\mathrm{Im}(c_l^k) & \mathrm{Re}(c_l^k)
\end{array}\right)_{1\leq k,l\leq n}\in\mathrm{GL}(2n,\mathbb R),$$
using the Cauchy-Riemann equations. We will calculate the determinant of these matrices, show that it is always $>0$, which is equivalent to $M^{\mathbb R}$ being orientable.
We move on to calculating the determinants of these. Consider the $\mathbb R$-algebra homomorphism
$$\rho:M_n(\mathbb C)\rightarrow M_{2n}(\mathbb R),~\left(c_l^k\right)_{1\leq k,l\leq n}\mapsto \left(\begin{array}{cc}
\mathrm{Re}(c_l^k) & -\mathrm{Im}(c_l^k) \\
\mathrm{Im}(c_l^k) & \mathrm{Re}(c_l^k)
\end{array}\right)_{1\leq k,l\leq n}$$
Since it is $\mathbb R$-linear and the spaces involved are finite dimensional, it is continuous. Also, being an algebra homomorphism, we have $\mathrm{det}~\rho(P^{-1}AP)=\mathrm{det}(\rho(P)^{-1}\rho(A)\rho(P))=\mathrm{det}(\rho(A))$. Finally, the diagonalizable matrices are dense in $M_n(\mathbb C)$, so we can restrict our calculations to diagonal matrices in $M_n(\mathbb C)$. For these, the calcuations are easy:
$$\mathrm{det}\big(\rho(\mathrm{Diag}(c_1,\dots,c_n))\big)=\mathrm{det}~\mathrm{Diag}\left(\left(\begin{array}{cc}
\mathrm{Re}(c_1) & -\mathrm{Im}(c_1) \\
\mathrm{Im}(c_1) & \mathrm{Re}(c_1)
\end{array}\right),\dots,\left(\begin{array}{cc}
\mathrm{Re}(c_n) & -\mathrm{Im}(c_n) \\
\mathrm{Im}(c_n) & \mathrm{Re}(c_n)
\end{array}\right)\right)\\=\prod_{i=1}^n\mathrm{det}\left(\begin{array}{cc}
\mathrm{Re}(c_i) & -\mathrm{Im}(c_i) \\
\mathrm{Im}(c_i) & \mathrm{Re}(c_i)
\end{array}\right)=\prod_{i=1}^n|c_i|^2=\left|\mathrm{det}~\mathrm{Diag}(c_1,\dots,c_n)\right|^2,$$
so we conclude that
$$\forall A\in M_n(\mathbb C),~\mathrm{det}~\rho(A)=|\mathrm{det}~A|^2$$
Finally, we can conclude that the transition functions for the charts $\phi^{\mathbb R}$ for $M^{\mathbb R}$ have positive determinants, thus the real underlying manifold $M^{\mathbb R}$ is orientable.
Yes, the second statement implies that the quotient (with quotient topology) is Hausdorff.
You can relax the hypotesis requiring that $g(N_1) \cap N_2 \neq \emptyset$ for at most a finite set of $g \in G$ (this is useful when checking the condition for example on complex tori).
The proof of this is a game of intersecting open subsets. Start with the quotient, recall what open subsets are, see what you need on the preimages such that the intersection of the images is empty.
The first statement is properness (in the sense of covering maps) of the action. It implies that the quotient map
$$\pi \colon X \twoheadrightarrow X/G$$
is a covering map and in particular a local homeomorphism.
Let $y = [x] \in X/G$, choose a lift $x \in X$, statement (1) says that there is
a $x \in U \subseteq X$ such that $g(U) \cap U \neq \emptyset$ implies $g = id$.
This in particular gives that if $g_1 \neq g_2$ then
$$g_1(U) \cap g_2(U) = \emptyset.$$
Let $[U] = \pi(U)$ be the image, which is open. Then
$$\pi^{-1}([U]) = \bigsqcup_{g \in G} g(U)$$
which shows that $\pi$ is a covering map. In particular the restriction
$$\pi_{|} \colon U \to [U]$$
is a homeomorphism, and you use those to push charts.
I have been a bit sketchy,
but you can find all this in any book on covering maps.
Best Answer
No, this is not always possible. Indeed, the one-point compactification of a Riemann surface may not even be a manifold. For instance, the one-point compactification of $\mathbb{C}\setminus\{0\}$ is a sphere with two points identified, which is not locally homeomorphic to $\mathbb{R}^2$ at that point.
If $S$ and $T$ are Riemann surfaces which happen to have one-point compactifications $S^*$ and $T^*$ which are Riemann surfaces, then any proper holomorphic map $f:S\to T$ does extend holomorphically to a map $S^*\to T^*$. By properness, we can extend $f$ continuously to a map $S^*\to T^*$, and this map is holomorphic at $\infty$ since the singularity must be removable (in coordinate charts at $\infty$, $f$ is bounded near the singularity).