Q: Exactly what information is contained in f*(OX)? Look at the definition. For any UinY open, f*(OX)(U) = OX(f-1(U)) = regular functions on f-1(U). So the information in f*(OX) is related to the sets in X of form f-1(U).
Cases where f*(OX) contains as little information about X as possible.
If X is irreducible and projective and f is constant, e.g. if Y is affine, then the only non empty set of form f-1(U) in X is X itself. In this case f*(OX) is a skyscraper sheaf with stalk k supported on the image point of f in Y. There is very little information here about X, but perhaps we do see that f is constant and that X is connected. More generally, if Z is a projective variety, Y is any variety, and X = ZxY, and f:ZxY→Y is the projection, then f-1(U) = ZxU, so an element of f*(OX)(U), i.e. a regular function on f-1(U), is determined by its restriction to {p}xU for any p in X, i.e. a regular function on UinY. Thus in this case we have f*(OX) = OY. Consequently in this case f*(OX) recovers Y, but contains no information at all about X.
In general, if f:X→Y is a projective morphism with every fiber connected, and Y is any normal variety, then f*(OX) = OY, so again f*(OX) contains little information about X. Recall that if X is a projective variety then every morphism out of X is a projective morphism, and more generally a projective morphism X→Y is one that factors via an isomorphism of X with a closed subvariety of PnxY, followed by the projection PnxY→Y. Suppose that f:X→Y is any projective morphism. Then the fibers f-1(y) over points y in Y are all finite unions of projective varieties. Therefore for any open set UinY containing the point y, the only regular functions in OX(f-1(U)) = f*(OX)(U) are constant on every connected component of the fiber f-1(y). Thus f*(OX) can contain little information about X and f, other than at most the connected components of the fibers. We shall see below that it contains exactly this information.
Cases where f*(OX) contains as much information about X as possible.
If f:X→Y is a map of affine varieties, then the global sections of f*(OX) determine X completely, since then H^0(Y,f*(OX)) = H^0(X,OX), and then X = specH^0(X,OX), is the unique affine variety with coordinate ring H^0(X,OX). The generalization of this case is that of any affine map f:X→Y, since then X can be recovered by patching together the analogous construction from H^0(U,f*(OX)) for affine open sets UinY. Thus X is completely determined by f*(OX) for any affine map f:X→Y, and this is essentially the only case. I.e. in general f*(OX) is always a quasi coherent OY algebra, and if we want it to determine a variety, as opposed to a "scheme", it is reasonable to assume for all UinY affine open, that f*(OX)(U) is a finitely generated k algebra, as well as an OY(U) algebra. We may call temporarily such an OY algebra "of finite type". Thus if f:X→Y is any morphism such that f*(OX) is of finite type, then the patching construction above yields not necessarily X, but a variety Z and an affine map h:Z→Y which factors via a map g:X→Z, where f = hog, and where g*(OX) = OZ. In particular then, we have f*(OX) = (hog)(OX) = h(g*(OX))= h*(OZ). So since h is affine, f*(OX) = h*(OZ) determines not X, but Z. (Kempf, section 6.5.)
The case of an arbitrary projective morphism.
Now when f:X→Y is any projective morphism, then f*(OX) is a coherent OY module, hence we get a factorization of f as hog:X→Z→Y, where h:Z→Y is affine, and where also h*(OZ) = f*(OX). Then h is not only an affine map, but since h*(OZ) is a coherent OY module, h is also a finite map. Moreover g:X→Z is also projective and since g*(OX) = OZ, it can be shown that the fibers of g are connected. Hence an arbitrary projective map f factors through a projective map g with connected fibers, followed by a finite map h. Thus in this case, the algebra f*(OX) determines exactly the finite part h:Z→Y of f, whose points over y are precisely the connected components of the fiber f-1(y).
One corollary of this is "Zariski's connectedness theorem". If f:X→Y is projective and birational, and Y is normal then f*(OX)= OY, and all fibers of f are connected, since in this case Z = Y in the Stein factorization described above. If we assume in addition that f is quasi finite, i.e. has finite fibers, then f is an isomorphism. More generally, if Y is normal and f:X→Y is any birational, quasi - finite, morphism, then f is an embedding onto an open subset of Y ("Zariski's 'main theorem' "). More generally still, any quasi finite morphism factors through an open embedding and a finite morphism.