My question refers to a step in the proof of Thm. 16.4.1 (page 424) in Ravi Vakil's script "Foundations of Algebraic Geometry":
Here the whole pdf:
math.stanford.edu/~vakil/216blog/FOAGdec3014public.pdf
Let $\pi:X \to \mathbb{P}^n _R$ a morphism of schemes, where $\mathbb{P}^n _R= Proj(R[x_0, x_1, ... , x_n]$. Obviously every global section $x_i \in \Gamma(X, \mathcal{O}_{\mathbb{P}^n _R}(1))$ can be interpet as morphism $x_i:\mathcal{O}_{\mathbb{P}^n _R} \to \mathcal{O}_{\mathbb{P}^n _R}(1)$, therefore $s_i := \pi^*x_i \in \Gamma(X, \pi^*(\mathcal{O}_{\mathbb{P}^n _R}(1)))$ is well defined.
Let restrict $\pi$ to $D_+(s_i) \to D_+(x_i)= Spec(R[x_0/x_i, x_1/x_i, ... , x_n/x_i]$. Since $D_+(x_i) \subset \mathbb{P}^n _R$ affine, the restricted morphism $\pi \vert_{D_+(s_i)}:D_+(s_i) \to D_+(x_i)$ is completely determined by $(\pi \vert_{D_+(s_i)})^ {\#}: R[x_0/x_i, x_1/x_i, ... , x_n/x_i] \to \Gamma(D_+(s_i), \mathcal{O}_X), x_j/x_i \to (\pi \vert_{D_+(s_i)})^ {\#}(x_j/x_i)$ on the "ring side".
Remark: Here $(\pi \vert_{D_+(s_i)})^ {\#}$ is the ring morphism on global sections of $\pi \vert_{D_+(s_i)}$.
My question is why $(\pi \vert_{D_+(s_i)})^ {\#}(x_j/x_i)= s_j/s_i$?
Think of what happens when $R$ is an algebraically closed field $k$. Then $\mathbb{P}^n$ is a set of equivalence classes $[a_0:\cdots:a_n] = \{(\lambda a_0, \dots,\lambda a_n):\lambda \in k^\times\}$. To give a morphism $\pi : X \to \mathbb{P}^n$ is to give $n + 1$ functions $s_0,\dots,s_n : X \to k$ such that
$$ \pi(x) = [s_0(x) : \dots :s_n(x)] $$
For this to be well-defined, $s_0,\dots,s_n$ should not have a common zero.
The section $x_i$ is the $i$-th coordinate of $\mathbb{P}^n$ and its pullback under $\pi$ is $s_i$, the $i$-th coordinate of $\pi(X)$. To make this all precise, we turn to the affine coordinate rings.
The affine open $D(x_i)$ is
$$ D(x_i) = \operatorname{Spec} k[x_0/x_i, \dots, x_n/x_i] = \{[a_0 : \dots : a_n] \in \mathbb{P}^n : a_i = 1\} $$
We have $\pi^{-1}(D(x_i)) = \{ x \in X : s_i(x) \ne 0 \} = D(s_i)$. The affine morphism we get is
$$ \pi|_{D(s_i)} : x \mapsto \left( \frac{s_0(x)}{s_i(x)}, \dots, \widehat{\frac{s_i(x)}{s_i(x)}} , \dots, \frac{s_n(x)}{s_i(x)} \right). $$
Thus,
$$ (\pi|_{D(s_i)})^\sharp(x_j/x_i) = s_j/s_i. $$
Recall: if you have a morhism of affine varieties $f : X \to \mathbb{A}^n$ given by $f(x) = (f_1(x),\dots,f_n(x))$ then the map of coordinate rings is given by $f^\sharp : k[y_1,\dots,y_n] \to k[X]$, $f^\sharp(y_j) = f_j$.
When $R$ is any ring, not necessarily an algebraically closed field, then you can no longer thing of $\mathbb{P}^n_R$ as a set of equivalence classes in this way. You can however think of $\mathbb{P}^n_R$ as $n + 1$ copies of $\mathbb{A}^n_R$ glued together appropriately.
Given a morphism $\pi : X \to \mathbb{P}^n_R$. The sections $s_i := \pi^* x_i$ generate the invertible sheaf $\mathscr L := \pi^* \mathcal O(1)$. The transition functions for $\mathcal O(1)$ are $x_j/x_i$ and these pullback under $\pi$ to the transition functions $\pi^*(x_j/x_i) = s_j/s_i$ on $\mathscr L$ (c.f. 16.3.7.(3)).
We know what $\pi^*$ does to $x_0,\dots,x_n$. When we localize (homogeneously) at $x_i$, we we have isn't just a module, it's a ring. However, the ring map is still determined by the module map and the module map sends $x_j/x_i$ to $s_j/s_i$. Hence the ring map sends $x_j/x_i$ to $s_j/s_i$.