Finite surjective morphism of normal varieties and Galois coverings Let $f:X\to Y$ be a finite surjective morphism of normal varieties. We say that $f$ is a Galois covering if the extension $k(X)/k(Y)$ is Galois.
I have the following questions:
1.- Why we need to assume normality of the varieties for the definition of a Galois covering?
2.- Is it true that a finite surjective morphism between normal varieties is always flat? 
3.- Is it true that $f$ is a Galois covering if and only if the group of automorphism $Aut(X/Y)$ acts transitively on all fibers of $f$?
 A: I'm just going to make a quick additional comment on #2 (just to add on to Timo Keller's explanations and Sasha's counter example).
Theorem (see Matsumura, page 179) If $R \subseteq S$ is a finite extension of Noetherian rings with $R$ regular and $S$ Cohen-Macaulay, then $S$ is a flat $R$ module.
So you just need $Y$ to be regular and $X$ to be Cohen-Macaulay (Matsumura actually has a more general statement which also works in the non-finite case).
On the other hand, if $Y$ is not regular then $X/Y$ is very rarely flat.  Indeed, since the map is finite, flat is the same as $f_* O_Y$ being a locally free $O_X$-module and we can ask how many (local) $O_X$-summands $f_* O_Y$ has as an $O_X$-module.
Fact  If $R \subseteq S$ is a finite extension of normal local rings with the same residue field which is etale in codimension 1 (ie, the sort of group quotient that Sasha wrote above), then $S$ has at most one $R$-summand as an $R$-module (in particular, it is nowhere near flat).  See the question I asked earlier: Number of free summands of finite local extensions
A: The simplest counterexample to question 2 is the map 
$$
X := \mathbb{A}^2 \to \mathbb{A}^2/\{ \pm 1 \} =: Y.
$$
Here $Y$ is a quadratic cone in $\mathbb{A}^3$. Both $X$ and $Y$ are normal, and the map is a Galois covering, but $X$ is not flat over $Y$.
A: (I assume by a covering, you mean a finite surjective étale morphism.)


*

*This is needed as in your definition of Galois covering you only look at the extension of the function fields.  Using the more general definition from [SGA1, Exp. V, Remarque 5.11], this works for any connected scheme.  For this, see also [Stacks Project], http://stacks.math.columbia.edu/download/pione.pdf, p. 6, after Lemma 3.7: $X/Y$ is Galois (this works for any Galois category with fibre functor $\mathrm{Fib}_\bar{y}$) iff $X$ is connected and $|\mathrm{Aut}(X/Y)| = \mathrm{Fib}_\bar{y}X$ iff $X$ is connected and $\mathrm{Aut}(X/Y)$ acts transitively on $\mathrm{Fib}_\bar{y}X$.  The étale fundamental group is defined in Definition 6.1 there.  In section 10, the fundamental group of normal schemes is considered.  See also http://websites.math.leidenuniv.nl/algebra/GSchemes.pdf, p. 41, 3.14 Galois objects.  For the question what happens in the non-normal case, see The étale fundamental group in the non-normal case.

*If $f: X \to Y$ is a finite surjective morphism of regular schemes (or more generally, if $Y$ is regular and $X$ is Cohen-Macaulay), it is flat, see [Liu], Remark 4.3.11, or [EGAIV${}_3$], p. 230, Proposition (15.4.2).

*I think this is true, but I do not have a reference. I would show that $\mathrm{Aut}(X/Y)$ acting transitively on the fibres is equivalent to the usual definition of a Galois cover from [SGA1], and then that this is equivalent to the generic fibre being a Galois cover. I think this follows from the fact that one has a surjection $G_{k(Y)} \twoheadrightarrow \pi_1^{\mathrm{\acute{e}t}}(Y,\bar{y})$ and under surjective group homomorphisms, normal subgroups correspond to normal subgroups.
