Theorem 3.2, Chapter 1, of Hartshorne's Algebraic Geometry
Solution 1:
Question: "Could somebody explain this statement in detail?"
Answer: Let $A:=A(Y)$ be the coordinate ring of $Y$, which is an integral domain. There is for every $V \subseteq Y \subseteq Y$ pair of open subsets, canonical maps $\rho_{Y,U}: A(Y) \rightarrow \mathcal{O}(U)$ commuting with the restriction map
$$\mathcal{O}(U) \rightarrow \mathcal{O}(V).$$
Here $\mathcal{O}(U)$ is the ring of regular functions on $U$. For any element $f \in\mathcal{O}(U)$ it follows the equivalence class $<U,f> \in K(Y)$ (this is the notation in HH, CH.I.3). In particular we get a canonical map
$$\rho: A(Y) \rightarrow K(Y)$$
defined by
$$\rho(s):=<Y,s> \in K(Y).$$
if $s\neq 0$ it follows $Z(s) \subsetneq Y$ is a strict closed subset and $U:=Y-Z(s) \subseteq Y$ is a non-empty open set. It follows $<U,1/s>\in K(Y)$ hence $\rho(s)\in K(Y)$ is invertible and there is an induced sequence of injections
$$A(Y) \rightarrow K(A(Y)) \rightarrow^{\eta} K(Y)$$
hence $K(A(Y)) \subseteq K(Y)$ is an inclusion. Given any element $<U,f>\in K(Y)$ and a point $p\in U$ there is an open subset $p \in U_p \subseteq U$ and regular functions $g,h\in A(Y)$ with $h \neq 0$ on $U_p$ and $<U,f> \cong <U_p,g/h>$ in $K(Y)$. It follows $g/h \in K(A(Y))$ and $\eta(g/h)=<U,f>.$ Hence the map $\eta: K(A(Y)) \rightarrow K(Y)$ is surjective. This implies the inclusion $K(A(Y)) \subseteq K(Y)$ is an isomorphism.