In this MO comment, Brian Conrad stresses that Zariski's main theorem is the engine facilitating the structure theory of smooth, unramified, and étale morphisms.
In the theory of smooth manifolds, the local structure theory of submersions and immersions seems to rely "only" on the inverse function theorem.
Geometrically, what is the role of ZMT in the structure theory of smooth, unramified, and étale morphisms? How does it encompass passage from the infinitesimal to the local, as the inverse function theorem does in the smooth category?
In the analytic theory one shows in the beginning that the ring of convergent power series is Henselian---in the meaning that, for morphism between such rings, quasi-finite implies finite.
In algebraic geometry this is not true for the local rings. The meaning of Zariski's main theorem is that a quasi-finite morphism can be regarded as an open immersion into some scheme which lies finite over the target of the morphism. Then it is easy to see that an unramified morphism can (locally) be imagined as an open subscheme which is a subscheme of the affine line over the target. Then it is somehow tautotogical that, after an étale base change, an unramified scheme becomes a closed immersion. For example, if the morphism is étale, then the base change is the tautological one by the given morphism.
We should look at an example. So consider the following.
Let$$f := a_nT^n + \ldots + a_0T^0 \in A[T]$$be a polynomial over a ring $A$. Set$$X := \text{Spec}\,A, \text{ } Y := V(f) - V(a_n) \subset \mathbb{A}_X^1.$$Then the canonical morphism $Y \to X$ is quasi-finite but not finite if $V(a_n)$ is not empty.
How do we get an embedding $Y \hookrightarrow Z$ of $Y$ into a scheme $Z$ which is finite over $X$?
Yes, introduce the $A$-algebra$$B := (A[T]/(f))_{a_n}.$$Then put$$C := \{b \in B : b \text{ integral over }A\}, \text{ }Z := \text{Spec}\,C.$$Then we have $B = C_{a_n}$, and hence we obtain an open immersion $Y \hookrightarrow Z$. Now we use the fact that $C$ is of finite type over $A$, and hence finite over $A$. Henceforth $Z \to X$ is finite.
Update: There is not much to visualize. Extending a quasi-finite morphism to a finite one is sort of a "completion" process. Also note that, locally over the base, finite morphisms can be handled quite well in terms of finite ring extensions, whereas a similar procedure is not possible for quasi-finite morphisms.