What prevents a cover to be Galois? Let $f:X\rightarrow Y$ be a ramified cover of Riemann surfaces or algebraic curves over $\mathbb{C}$. My question is can one in terms of the ramification data of $f$, determine whether the cover is Galois or not. I am Specially interested in the following question: if $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are both ramified Galois covers of curves, when  $g\circ f:X\rightarrow Z$ is (or isn't) a Galois cover?
 A: The ramification data doesn't generally suffice to determine whether the cover is Galois.  If $f:X\to Y$ and $g:Y\to Z$ are ramified Galois covers, then usually $g\circ f$ won't be Galois, for instance because there will usually be a point $z$ of $Z$ which has two preimages under $g\circ f$ having different ramification indices.
A: There are many automorphisms of $Y$ over $Z$. One can pull back the cover $X$ along any of these automorphisms of $Y$, producing a new cover of $Y$. If the composition is Galois, then these new covers are in fact isomorphic to the original cover - this is because the map $Gal(X|Z) \to Gal(Y|Z)$ is surjective. Taking a lift of an element of $Gal(Y|Z)$ produces an isomorphism between the original cover and the pullback.
The converse is also true. If such an isomorphism exist, they give us enough elements of the automorphism group of $X$ over $Z$ that, combined with the automorphisms of $X$ over $Y$, the cover is Galois!
So this explains why the two preimages with different ramification data will prevent the cover from being Galois. Because then the pullback of the cover along an automorphism which permutes those two preimages will not be isomorphic to the original - it will have different ramification data.
A: I assume that the surfaces are connected. Composition of two regular covering maps need not be regular even if the coverings are unbranched: A (usual) covering map is regular iff it is defined by a normal subgroup of the fundamental group. If you have a composition
$$
X \stackrel{f}{\to} Y \stackrel{g}{\to} Z
$$
of regular covers then $\pi_1(X)\triangleleft \pi_1(Y)$ and 
$\pi_1(Y)\triangleleft \pi_1(Z)$. Hence,  $\pi_1(X)$ is a subnormal subgroup of 
$\pi_1(Z)$. Subnormal subgroups need not be normal. However, if, say, 
$\pi_1(X)< \pi_1(Y)$ is a characteristic subgroup then it will be normal in $\pi_1(Z)$. This is the simplest condition I know to ensure that composition of regular covering maps is again regular. The same applies to ramified covers once you remove branch points and their preimages.  
