In harshorne, proposition 3.5 we are trying to establish a bijection between the set of morphisms of varieties and homomorphims of k-algebras. $$ \alpha: \operatorname{Hom}(X, Y) \simeq \operatorname{Hom}(A(Y), \mathcal{O}(X)) $$ where $X$ is a variety and $Y$ an affine one. $A(Y)$ is the coordinates ring of $Y$ and $\mathcal{O}(X)$ is the ring of regular functions on $X$
First we construct $\alpha$ in a natural way this part i get it. Next we try to construct a morphism of varieties frome a given homomorphism of $k$-algebras :
let $h$ a morphism of $k$-algebras $h: A(Y) \rightarrow\mathcal{O}(X)$. Let $\overline{x_i}$ be the class of $x_i$ in $A(Y)$,and consider the element $\xi_{i}=h\left(\bar{x}_{i}\right) \in \mathcal{O}(X)$. we define the mapping $$\psi: X \rightarrow \mathbf{A}^{n},\quad\psi(P)=\left(\xi_{1}(P), \ldots, \xi_{n}(P)\right)\ \ \forall P \in X$$ Let show that the image of $\psi$ is contained in $Y$. Since $Y=Z(I(Y))$, it is sufficient to show that for any $P$ in $X$ and any $f \in I(Y), f(\psi(P))=0$. But $$f(\psi(P))=f\left(\xi_{1}(P), \ldots, \xi_{n}(P)\right)$$ Now $f$ is a polynomial and $h$ a homomorphism of $k$-algebras, so we have $$f\left(\xi_{1}(P), \ldots, \xi_{n}(P)\right)=h\left(f\left(\bar{x}_{1}, \ldots, \bar{x}_{n}\right)\right)(P)=0$$
what i dont undestand is the two last equalities. How we reversed $f$ and $h$ !! and $f$ is supposed to act on points not classes of polynomials and why it equals zero.
Question: "How we reversed f and h !! and f is supposed to act on points not classes of polynomials and why it equals zero."
Answer: There is no need to speak of variables $x_i$ and generators for $A(Y)$. Let $A(Y):=k[x_1,..,x_n]/I$ and assume you are given a map of $k$-algebras
$$f: A(Y) \rightarrow \Gamma(X,\mathcal{O}_X).$$
Let $U:=Spec(A) \subseteq X$ be an open affine subscheme. There is an equality $\Gamma(U, \mathcal{O}_X)=A$ and you get a restriction map
$$\rho_U: \Gamma(X, \mathcal{O}_X) \rightarrow \Gamma(U, \mathcal{O}_X)=A$$
and a composed map $ f_A:=\rho_U \circ f: A(Y) \rightarrow A$ and a corresponding map of affine schemes
$$F_U: Spec(A):=U \rightarrow Y:=Spec(A(Y)).$$
Hence for any open affine subscheme $U\subseteq X$ you get a canonical map $F_U: U \rightarrow Y$ and it follows (since all maps are canonical, induced by the restriction map $\rho_U$) that these maps glue to a well defined map $F: X \rightarrow Y$: Given any open affine scheme $V:=Spec(B)$ and any open affine subscheme $W:=Spec(R) \subseteq U \cap V$ you get canonical maps
$$F_U: U \rightarrow Y, F_W:W \rightarrow Y$$
and since "all diagrams commute", it follows
$$F_U \circ i_U = F_W = F_V \circ i_V$$
where $i_U: W \rightarrow U$ is the inclusion $W \subseteq U$. Hence the maps $F_U$ glue to a map $F: X \rightarrow Y$.