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There are many formulations of Zariski's main theorem, and Mumford's Red Book gives a very nice description of some of them, and of their interrelations.

It is worthwhile remembering that it was in fact Zariski who first proved a version of his main theorem. Zariski was a geometer first and foremost (although he was also one of the greatest ever commutative algebraists), and so it is reasonable to look for the geometric content in this result.

In fact Zariski first proved his main theorem before he developed his theory of formal functions (which was his method for proving connectedness theorems, and in its modern cohomlogical reformulation by Grothendieck remains the basic method for proving connectedness statements, as in Francesco Polizzi's answer). Zariski's original version of his theorem stated that if the preimage of a point on a normal variety under a birational map contains an isolated point, then the birational map is in fact an isomorphism in a n.h. of that point.

As Mumford explains, what this result and the later variations have in common is that a variety is unibranch at a normal point, i.e. there is only one branch of the variety passing through such a point. Thus, if we blow the variety up in some way, we can might be able to increase the dimension at this point (in the sense that we might be able to replace $y$ by something higher dimensional), but we cannot break the variety apart there.

Grothendieck's formulation is very natural: it states that is we have a quasi-finite morphism, we can always compactify it (compactification to be understood in a relative sense) to a finite morphism. To see how this implies Zariski's original result, just observe that if $f:X \to Y$ is birational with $Y$ normal, and $x$ is an isolated point in $f^{-1}(y)$, then we can choose a n.h. $U$ of $x$ over which $f$ is quasi-finite, which then compactifies to a finite morphism. But since $Y$ is normal, it any finite birational map to $Y$ must be an isomorphism, and so $U$ must be emdedding embedding into $Y$ via an open immersion. In short, $f$ is an isomorphism between a n.h. of $x$ and a n.h. of $y$.
show/hide this revision's text 1

There are many formulations of Zariski's main theorem, and Mumford's Red Book gives a very nice description of some of them, and of their interrelations.

It is worthwhile remembering that it was in fact Zariski who first proved a version of his main theorem. Zariski was a geometer first and foremost (although he was also one of the greatest ever commutative algebraists), and so it is reasonable to look for the geometric content in this result.

In fact Zariski first proved his main theorem before he developed his theory of formal functions (which was his method for proving connectedness theorems, and in its modern cohomlogical reformulation by Grothendieck remains the basic method for proving connectedness statements, as in Francesco Polizzi's answer). Zariski's original version of his theorem stated that if the preimage of a point on a normal variety under a birational map contains an isolated point, then the birational map is in fact an isomorphism in a n.h. of that point.

As Mumford explains, what this result and the later variations have in common is that a variety is unibranch at a normal point, i.e. there is only one branch of the variety passing through such a point. Thus, if we blow the variety up in some way, we can increase the dimension at this point, but we cannot break the variety apart there.

Grothendieck's formulation is very natural: it states that is we have a quasi-finite morphism, we can always compactify it (compactification to be understood in a relative sense) to a finite morphism. To see how this implies Zariski's original result, just observe that if $f:X \to Y$ is birational with $Y$ normal, and $x$ is an isolated point in $f^{-1}(y)$, then we can choose a n.h. $U$ of $x$ over which $f$ is quasi-finite, which then compactifies to a finite morphism. But since $Y$ is normal, it any finite birational map to $Y$ must be an isomorphism, and so $U$ must be emdedding into $Y$ via an open immersion. In short, $f$ is an isomorphism between a n.h. of $x$ and a n.h. of $y$.