The following is an exercise in Liu (exercise 3.3.13(b), page 111):
Let $f$ be a dominant rational map $X\dashrightarrow Y$ where $X$ and $Y$ are integral schemes of finite type over a locally Noetherian scheme $S$. Show that a point $x\in X$ lies in the domain of definition if and only if there is a point $y\in Y$ such that $\mathscr{O}_{Y,y}$ is dominated by $\mathscr{O}_{X,x}$ under the field extension $K(Y)\to K(X)$.
This reduces in a straightforward way to a the affine case, where it becomes a statement in commutative algebra. The proof of that statement, as far as I can tell, is just as straightforward. The thing that has me worried is that I believe this holds with no Noetherianness condition on $S$ ($S$ can be an arbitrary scheme) and no finite type condition on $X$. Those conditions, of course, are pretty mild, so it's not surprising that Liu would include them anyways, but I want to make sure that I'm not missing a major obstruction.
Here's my statement/idea of proof of that local case:
Let $A$ and $B$ be $R$-algebras, with $A$ finite type, and suppose we have a homomorphism $f:A\to B_\beta$ for some $\beta\in B$. Then for any prime $\mathfrak p$ of B such that $f(A)\subseteq B_\beta \cap B_\mathfrak p$, there is $\beta'\notin \mathfrak p$ such that $f(A)\subseteq B_\beta\cap B_{\beta'}$.
The proof is shorter than the statement. Choose $R$-generators $x_1,\dots,x_n$ for $A$. Then $f$ is given by $f(x_i) = f_i/\beta^n$. The condition $f(A)\subseteq B_\mathfrak p$ means that we have $t_i\notin\mathfrak p$ so that $t_if(x_i)\in B$. If we let $\beta'=\prod t_i$, then $f(A)\subseteq B_\mathfrak p$, Q.E.D.
A note to relate this to global case: having an affine patch of $Y$ whose coordinate ring maps into the local ring is just as good as having a point whose local ring is dominated: we can just pull back the maximal ideal of the local ring, and then localize.
Certainly knowing $A$ can be described by only finitely many relations on the $x_i$ comes only if $S$ is locally Noetherian, but is there any reason we benefit from that? And is there any reason why we would need some sort of finiteness conditions on $X$?
In the June 11, 2013 edition of Vakil's FOAG, proposition 6.5.7 is a closely related (read: equivalent) statement we have this statement without f.t. assumptions on $X$, but with $S$ replaced by a field, rather than an arbitrary locally Noetherian scheme.
Seeing as I didn't actually have a particular point of worry in the proof, but rather in the statement, seeing this alternate statement is enough to quell my concerns. (Not to mention the fact that Vakil's proof is essentially the same as the one I gave.)