At page $59$ in Görtz & Wedhorn 's "Algebraic Geometry 1" book, there is the following text:
"Let $\varphi : A \rightarrow B$ be a homomorphism of rings and set $X = \operatorname{Spec} B$ and $Y = \operatorname{Spec} A$. Let $^a\varphi : \operatorname{Spec} B \rightarrow \operatorname{Spec} A$ be the associated continuous map. We will now define a morphism $(f,f^b):X \rightarrow Y$ of locally ringed space and such that $f = ^a\varphi$ and $f^b_Y:A=\mathscr O_Y(Y) \rightarrow (f_*\mathscr O_X(Y))=B$ equals $\varphi$.
Set $f = ^a\varphi$, for $s \in A$, We have $f^{-1}(D(s))=A_s \rightarrow B_{\varphi(s)}= (f_*\mathscr O_X)(D(s))$ is the homomorphism induced by $\varphi$. This ring homomorphism is compatible with restrictions to principal open subsets $D(t) \subset D(s)$. As the principal open subsets form a basis of the topology, this defines a homomorphism $f^b:\mathscr O_Y \rightarrow f_*\mathscr O_X$ of sheaves of rings.
For $x \in X$, the homomorphism $f^\#_x: \mathscr O_{Y,f(x)} = A_{\varphi^{-1}(p_x)} \rightarrow B_{p_x} = \mathscr O_{X,x}$ is the homomorphism induced by $\varphi$ and in particular it is a local ring homomorphism. This finishes the definition of $(f,f^b)$."
I understand the first two paragraphs. So we have a special homomorphism $(f,f^b)$ of locally ringed spaces $(X,\mathscr O_X)$ and $(Y,\mathscr O_Y)$. This $(f,f^b)$ is special enough such that for $s \in A$, $f^b_{D(s)}:f^{-1}(D(s))=A_s \rightarrow B_{\varphi(s)}= (f_*\mathscr O_X)(D(s))$ is the homomorphism induced by $\varphi$, that is $f^b_{D(s)}(\frac {a} {s^n}) = \frac {\varphi(a)} {\varphi(s)^n}$, and $f=a^\varphi$. Now we can say that $(f,f^b)$ is fixed because for $s\in A$,$D(s)$ is the principal open subsets so that $f^b$ is fixed. And $f^b$ is fixed so is $f^\#$ since $f^b$ and $f^\#$ have an one-to-one correspondence.
Now we have $f^\#$ being fixed, so for $x\in X$, the local ring homomorphism defined as $f^\#_x$ is fixed.
Then I have trouble understanding the third paragraph which says "For $x \in X$, the homomorphism $f^\#_x: \mathscr O_{Y,f(x)} = A_{\varphi^{-1}(p_x)} \rightarrow B_{p_x} = \mathscr O_{X,x}$ is the homomorphism induced by $\varphi$".
That means $f^\#_x(\frac {a} {q^n}) = \frac {\varphi(a)} {\varphi(q)^n}$.
But it is not at all obvious, right? So why is $f^\#_x$ being the homomorphism such that $f^\#_x(\frac {a} {q^n}) = \frac {\varphi(a)} {\varphi(q)^n}$?
Consider an arbitrary but fixed element $x\in A_{\varphi^{-1}({\mathfrak q})}$: It is then of the form $\tfrac{a}{b}$ for $b\in\varphi^{-1}(B\setminus{\mathfrak q})$, hence can also be viewed as the image of the 'same' element $\tfrac{a}{b}$ interpreted as an element of $A_b$, under the canonical homomorphism $A_b\to A_{\varphi^{-1}({\mathfrak q})}$. Next, by naturality, the composition $A_b\to A_{\varphi^{-1}({\mathfrak q})}\to B_{\mathfrak q}$ is the same as the composition $A_b\to B_{\varphi(b)}\to B_{\mathfrak q}$. Finally, you have already identified $A_b\to B_{\varphi(b)}$ in the latter composition as being the canonical map. Can you fill in the details here?