Function Field of Variety and Scheme Let $V\subseteq \mathbb{A}^n_k$ be a closed irreducible algebraic set ("affine variety") over a closed field $k$.
Construct the topological space $X$ consisting of all closed irreducible subsets of $V$ (it is given the Zariski topology). Note that we have a continous map $f:V\to X$ given by $f(x) = \overline{\{x\}}$. 
Let $\mathcal{O}_V$ be the sheaf of regular functions over $k$ on $V$. Push the sheaf, the ringed space $(X,f_*\mathcal{O}_V)$ is an affine scheme. 
Is the following true? Given $V'\in X$, $k(V')\simeq K(V')$. Where $k(V') = $ residue field of the scheme $X$ at $V'$, and $K(V') = $ the field of rational functions of $V'$. 

I think exactly how to answer your question may depend on the definitions you are using.
To set definitions:
On the side of varieties: $K(V') = frac(k[X]/I(V'))$, where $k[X]$ is the coordinate ring of $X$. (And $V'$ is serving double duty, here as an algebraic set.)
The residue field of the scheme $X$ at $V'$ is the residue field of the ring $k[X]$ localized at the prime $V'$.
So the commutative algebra side of your question is: 
Given a ring $R$ and a prime ideal $P$, is $frac(R/P) \cong R_P / PR_P$. The answer is yes: localization commutes with quotients ("localization is exact"), and localizing $R/P$ at $P$ is the same as constructing its field of fractions, because we are inverting all nonzero elements.
