Reading these excellent answers made me realize I did not understand the ZMT in an intuitive way, and helped me very much get at least more intuition for it than I had. I want to try to summarize what I have learned from thinking about the other answers here, in a way that is not too technical and hopefully intuitive. I do not consider the proofs, but only the intuition contained in the statements.

The first step is necessarily a little technical since it involves the definition of normality.

1) If X is a normal affine variety, i.e. if its affine ring has no non trivial module - finite extension within its quotient field, then it follows from this definition, that X accepts no non trivial finite birational morphism. Thus X is normal if and only if every finite birational morphism Y→X is an isomorphism.

2) ZMT then implies that every such normal variety is unibranch. I.e. if X has more than one branch at any point, then X must accept a non trivial finite birational map Y→X. The geometric intuition here is that it should be possible, by a finite morphism, to separate the branches of any variety that is not unibranch. This may have been suggested to Zariski by examples such as the projection cited by Francesco. Thus ZMT establishes that the only way for a variety to possess more than one branch is for it to be the target of a map similar to a “projection”. (Mumford's nice topological and power series statements are just alternate ways to say "unibranch".)

3) As Sandor pointed out, then the connectedness theorem follows naturally for a normal variety X, since if Y→X were finite, birational, and some fiber f^(-1)(p) were disconnected, then X should have at least two branches at p, hence X should not be normal.

4) The next natural piece of intuition is Grothendieck’s theorem that all quasi-finite morphisms can be completed to finite ones, as Matt and Akhil observe. Thus the only way to get a quasi - finite morphism is to restrict a finite morphism to an open set, a very intuitive geometric statement. In particular, since there are by definition no non - trivial finite birational morphisms to X, there cannot be any non trivial quasi- finite birational ones either; i.e. every such morphism Y→X is an open immersion.

5) As a consequence, a birational morphism Y→X which is not trivial, i.e. not an open immersion, cannot be quasi – finite, hence must have a positive dimensional fiber f^(-1)(p).

So for me at least, these various statements have all become geometrically natural, thanks to contemplating the other answers, admittedly in a naïve way.

arbitraryfinite birational morphism. For instance the normalization of a singular projective curve. Let $X=X'$. This seems to satisfy your condition. So, what is that statement without normality of $Y$? – Sándor Kovács Oct 21 '11 at 3:11