Let $A$ be a ring and let $G$ be a finite group with a morphism $G\to \mathrm{Aut}(A)$. Note that $G$ acts from the right on $X=\mathrm{Spec} A$. Consider the morphism $\pi:X\longrightarrow Y$, with $Y=\mathrm{Spec} A^G$, induced by the injection $A^G \subset A$.
Using that the category of affine schemes is (anti-)-equivalent to the category of rings, it is easy to see that $\pi:X\longrightarrow Y$ is the quotient for the action of $G$ in the category of affine schemes: every $G$-invariant morphism $f:X\to Z$ with $Z$ affine factors uniquely through $\pi$. In fact, this can be easily deduced from the following obvious statement: any ring morphism $\varphi:R\to A$ such that for all $g$ in $G$ we have $g\varphi = \varphi$ factors uniquely through $A^G$.
So as you know, $\pi:X\to Y$ is the quotient for the action of $G$ in the category of schemes. But it's actually even better than that! The morphism $\pi:X\to Y$ is the quotient for the action of $G$ in the category of locally ringed spaces. Let me sketch how you can prove this.
Firstly, show that $\pi:X\to Y$ is the set-theoretical quotient map, i.e., $\pi$ induces a bijection between the $G$-orbits of $X$ and $Y$. This is the hard part. For example,
given a prime ideal $\mathfrak{p}$ in $B=A^G$, you will need to show that $B_{\mathfrak{p}} = (A_{\mathfrak{p}})^G$. (The natural map $B_{\mathfrak{p}} \to A_{\mathfrak{p}}$ factors uniquely through a map $B_{\mathfrak{p}} \to (A_{\mathfrak{p}})^G$. It suffices to show that this map is surjective. Use the definition of the stalk and write it out.) Also, to conclude, suppose that we have two distinct orbits $x_1 G$ and $x_2 G$ of primes lying over $\mathfrak{p}$. Then, applying the Chinese remainder theorem, you will that this is impossible.
So suppose that we have shown that $\pi:X\to Y$ is the set-theoretical quotient map. Then, it is easy to see that $Y$ has the quotient topology. In fact, $\pi$ is continuous and closed.
Now, to finish, you look at the natural morphism of sheaves $ \mathcal{O}_Y\to (\pi_ast \mathcal{O}_X)^G$. This is an isomorphism. (You can check this is on the principal opens $D(b)$ of $Y$, with $b\in B$. But then this boils down to showing that $B_{b} \to (A_b)^G$ is an isomorphism. We already showed this above.)
So this is a really rough sketch of how I recall the proof. Let me know if you need some more details.
Best Answer
First some good news: if a finite group $G$ acts on a projective variety $X$, nice or not, then the quotient always exists, so that you do not need to assume existence.
Another pleasant result: if $X$ is normal, so will be $X/G$. This explains your observation that a quotient of a smooth curve is smooth.
Even in higher dimensions, it can happen that the quotient $X/G$ of the smooth variety $X$ is smooth even though the action of $G$ has fixed points.
An interesting example is the action of the symmetric group $S_n$ on $X=(\mathbb P^1_k)^n$. The quotient variety is just $\mathbb P^n_k$. This is a projective geometric vision of the fundamental theorem on symmetric polynomials.
However you can also have situations where the quotient acquires singularities.
The simplest example is to divide out the (affine) variety $\mathbb A^2_k$ by the action of the two-element group $G=\lbrace I,s\rbrace$, where $s$ acts as $s\ast (x,y)=(-x,-y)$. The morphism $\mathbb A^2_k \to \mathbb A^3_k$ given by $u=x^2, v=xy, w=y^2$ descends to an embedding $X/G \stackrel {\sim} {\to} V \subset \mathbb A^3_k$, where the cone $V$ is given by the equation $v^2=uw$.
(Incidentally, this is a nice way to prove that $V$ is normal, by invoking the "pleasant result" above!)
The nature of the singularities (or absence thereof) is a whole industry in algebraic geometry: the key-worde are "rational singularities", "Du Val singularities",...
Two nice introductions to quotient varieties : Shafarevich's book Basic Algebraic Geometry and Mumford's Abelian Varieties.
The definitive reference is supposed to be Mumford-Fogarty's Geometric Invariant Theory, but I haven't read it.