Let $S:=Spec(A)$ with $G \subseteq Aut_{k-alg}(A)$ a finite subgroup of the group of $k$-algebra automorphisms of $A$, with $A^G$ the invariant ring. It follows the ring extension $A^G \subseteq A$ is an integral extension and you get a canonical map
$$\pi: S \rightarrow S/G:=Spec(A^G).$$
If $A^G$ is generated as a $k$-algebra by a finite set of elements $g_1,..,g_l \subseteq A$ you get a canonical surjective map of $k$-algebras
$$\phi: k[y_1,..,y_l] \rightarrow A^G$$
defined by $\phi(y_i):=g_i$, and an isomorphism $A^G \cong k[y_i]/I$ where $I:=ker(\phi)$ is the kernel of $\phi$. Hence $S/G \subseteq \mathbb{A}^l_k$ is an affine algebraic variety which may be singular. There are explicit examples in the litterature where such quotients are constructed (singular and non-singular quotients) and you find them at mathscinet or the zentralblatt search site. Similar constructions exist for positive dimensional groups. The "smoothness" of the affine variety $S/G$ depends on the ideal $I$ of relations between the generators $g_i$, and I do not believe there is a general criteria for this. If the ring $A$ has an $\mathbb{N}$-grading and the action of $G$ has the property that the ideal $I$ is a homogeneous ideal, the quotient $S/G$ may have singularities.
Question: "I could not find some explicit examples where smoothess of quotients is verified."
Answer: If $G:=SL(V)$ where $V$ is an $n$-dimensional complex vector space and $W \subseteq V$ is a $d$-dimensional subspace with $P \subseteq G$ the subgroup fixing $W$ it follows the quotient $G/P \cong \mathbb{G}(d,V)$ is the grassmannian variety of $d$-planes in $V$ and $G/P$ is a smooth algebraic variety. More generally if $W$ is a finite dimensional vector space over a field $k$ and your action satisfies $End(W)^{ss}=G$ for some closed sub-group $G \subseteq GL(W)$, it follows the quotient $G/H$ is a smooth quasi projective variety of finite type over $k$. You find a similar question here:
Quotient of a Lie algebra by a subalgebra - what is it?
If you look at Borel's book "Linear algebraic groups" you find in Lemma 1.8 (the closed orbit lemma) the full argument. Essentially if an algebraic group $G$ acts on a variety $V$ then each orbit $Gx$ is smooth. For simplicity: Given any pair of points $y,z \in Gx$ there is an element $g\in G$ with $gz=y$ since $y,z$ is in the same orbit. Since $g: Gx \rightarrow Gx$ is an automorphism of $Gx$ it follows the local rings $\mathcal{O}_{Gx,y}\cong \mathcal{O}_{Gx,z}$ are isomorphic.
Hence any two points $y,z$ in the orbit have isomorphic local rings. Generically $Gx$ is smooth and hence it follows $Gx$ is smooth at any point $y$. If a quotient $G/H$ exists with a left $G$-action, it follows any two points $x,y\in G/H$ will have isomorphic local rings
$$\mathcal{O}_{G/H,x}\cong \mathcal{O}_{G/H,y}$$
hence $G/H$ is smooth at any point. The existence of the quotient $G/H$ is proved in Theorem 6.8 in the Borel book. If $H\times F \rightarrow F$ is a left $H$-action where $F$ is smooth, you may construct a locally trivial fibration
with fiber $F$
$$ \pi:G\times F/H \rightarrow G/H,$$
and $G\times F/H$ is smooth.
In general if $\sigma: G\times X \rightarrow X$ is an action with one orbit $Gx$ it follows $X$ is smooth by the same argument.
Note: The "stacks project" is constructing "quotients" in a more general setting, you may be aware of this site.
