Q: Exactly what information is contained in $f_*\mathscr O_X$? Look at the definition. For any $U\subseteq Y$ open, $f_*\mathscr O_X(U) = \mathscr O_X(f^{-1}(U))$ = regular functions on $f^{-1}(U)$. So the information in $f_*\mathscr O_X$ is related to the sets in $X$ of form $f^{-1}(U)$.
Cases where $f_*\mathscr O_X$ 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 $f_*\mathscr O_X$ 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 $X = Z\times Y$, and $f:Z\times Y\to Y$ is the projection, then $f^{-1}(U) = Z\times U$, so an element of $f_*\mathscr O_X(U)$, i.e. a regular function on $f^{-1}(U)$, is determined by its restriction to $\{p\}\times U$ for any $p\in X$, i.e., a regular function on $U$ in $Y$. Thus in this case we have $f_*\mathscr O_X = \mathscr O_Y$. Consequently in this case $f_*\mathscr O_X$ recovers $Y$, but contains no information at all about $X$.
In general, if $f:X\to Y$ is a projective morphism with every fiber connected, and $Y$ is any normal variety, then $f_*\mathscr O_X = \mathscr O_Y$, so again $f_*\mathscr O_X$ 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 $X\to Y$ is one that factors via an isomorphism of X with a closed subvariety of $\mathbb P^n\times Y$, followed by the projection $\mathbb P^n\times Y\to Y$. Suppose that $f:X\to Y$ 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 $U\subseteq Y$ containing the point $y$, the only regular functions in $\mathscr O_X(f^{-1}(U)) = f_*\mathscr O_X(U)$ are constant on every connected component of the fiber $f^{-1}(y)$. Thus $f_*\mathscr O_X$ 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 $f_*\mathscr O_X$ contains as much information about $X$ as possible.
If $f:X\to Y$ is a map of affine varieties, then the global sections of $f_*\mathscr O_X$ determine $X$ completely, since then $H^0(Y,f_*\mathscr O_X) = H^0(X,\mathscr O_X)$, and then $X = \mathrm{Spec}h^0(X,\mathscr O_X)$, is the unique affine variety with coordinate ring $H^0(X,\mathscr O_X)$. The generalization of this case is that of any affine map $f:X\to Y$, since then $X$ can be recovered by patching together the analogous construction from $H^0(U,f_*\mathscr O_X)$ for affine open sets $U\subseteq Y$. Thus $X$ is completely determined by $f_*\mathscr O_X$ for any affine map $f:X\to Y$, and this is essentially the only case. I.e. in general $f_*\mathscr O_X$ is always a quasi coherent $\mathscr O_Y$ algebra, and if we want it to determine a variety, as opposed to a "scheme", it is reasonable to assume for all $U\subseteq Y$ affine open, that $f_*\mathscr O_X(U)$ is a finitely generated k algebra, as well as an $\mathscr O_Y(U)$ algebra. We may call temporarily such an $\mathscr O_Y$ algebra "of finite type". Thus if $f:X\to Y$ is any morphism such that $f_*\mathscr O_X$ is of finite type, then the patching construction above yields not necessarily $X$, but a variety $Z$ and an affine map $h:Z\to Y$ which factors via a map $g:X\to Z$, where $f = h\circ g$, and where $g_*(\mathscr O_X) = \mathscr O_Z$. In particular then, we have $f_*\mathscr O_X = (h\circ g)_*(\mathscr O_X) = h_*(g_*(\mathscr O_X))= h_*(\mathscr O_Z)$. So since $h$ is affine, $f_*\mathscr O_X = h_*(\mathscr O_Z)$ determines not $X$, but $Z$. (Kempf, section 6.5.)
The case of an arbitrary projective morphism.
Now when $f:X\to Y$ is any projective morphism, then $f_*\mathscr O_X$ is a coherent $\mathscr O_Y$-module, hence we get a factorization of $f$ as $h\circ g:X\to Z\to Y$, where $h:Z\to Y$ is affine, and where also $h_*(\mathscr O_Z) = f_*\mathscr O_X$. Then $h$ is not only an affine map, but since $h_*(\mathscr O_Z)$ is a coherent $\mathscr O_Y$-module, $h$ is also a finite map. Moreover $g:X\to Z$ is also projective and since $g_*(\mathscr O_X) = \mathscr O_Z$, 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 $f_*\mathscr O_X$ determines exactly the finite part $h:Z\to Y$ 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 $f:X\to Y$ is projective and birational, and $Y$ is normal then $f_*\mathscr O_X= \mathscr O_Y$, 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 $f:X\to Y$ 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.
