I have a few questions concerning page 28 of this book.
Q1: In the definition of $\mathcal{O}_{\mathbb{P}^n(k)}(U)$ we have to check if $f_{\mid U\cap U_i}\in \mathcal{O}_{U_i}(U\cap U_i)$. We did not define what $\mathcal{O}_{U_i}$ is. I guess we take the isomorphism $U_i\ \overset{\sim}{\to} \mathbb{A}^n(k)$ and identify in that way $\mathcal{O}_{U_i}$ with $\mathcal{O}_{\mathbb{A}^n(k)}$.
A1: Sorry, I just saw that it is actually defined in exactly that way directly after mentioning the isomorphism $U_i\ \overset{\sim}{\to} \mathbb{A}^n(k)$.
Q2: The book says that we still have to see that $\mathcal{O}_{\mathbb{P}^n(k)\mid U_i}=\mathcal{O}_{U_i}$. Isn't that almost done by definition of $\mathcal{O}_{\mathbb{P}^n(k)}$? In the definition of $\mathcal{O}_{\mathbb{P}^n(k)}$ we want $f_{\mid U\cap U_i}\in \mathcal{O}_{U_i}(U\cap U_i)$ for all $i$. If we changed the definition slightly by just demanding $f_{\mid U\cap U_i}\in \mathcal{O}_{U_i}(U\cap U_i)$ only for our fixed $i$, would then $\mathcal{O}_{\mathbb{P}^n(k)\mid U_i}=\mathcal{O}_{U_i}$ be trivial by definition of $\mathcal{O}_{\mathbb{P}^n(k)\mid U_i}$ (Def. 1.36, p. 20)?
Q3: First part of the proof of Prop. 1.59: The book says "Applying the inverse of (1.19.2) yields the desired form of $f$". Clearly the inverse has the wanted form. But the connection between $f$ and the inverse is vague.
Q4: Cor. 1.60: How follows the first assertion from the proof of Prop. 1.59? We somehow have to see that Def. 1.35.2 holds in both directions (an isomorphism is a morphism in both directions). Continuity is clear, that was already mentioned before.
A2: Suppose that $U \subseteq U_i$. We clearly have $\mathcal{O}_{\mathbb{P}^n(k)|U_i}(U) \subseteq \mathcal{O}_{U_i}(U)$. In order to deduce the opposite inclusion $ \mathcal{O}_{U_i}(U) \subseteq \mathcal{O}_{\mathbb{P}^n(k)|U_i}(U)$, we have to show, in view of the definition of $\mathcal{O}_{\mathbb{P}^n(k)}(U) = \mathcal{O}_{\mathbb{P}^n(k)|U_i}(U)$, that for each element $f\in \mathcal{O}_{U_i}(U)$ and each $j\neq i$, we have $f_{|U\cap U_j} \in \mathcal{O}_{U_j}(U\cap U_j)$. As suggested by Prop. 1.59, we can do this as follows. Locally, around any given point in $U$, we take some $V \subseteq U$ on which $f$ can be represented by a fraction $p/q$ of polynomials in $n$ variables, with $q$ non-vanishing on $V$. First, we homogenise $p$ and $q$ to express $f$ as a quotient of polynomials $g$ and $h$ of the same degree in $(n+1)$ variables. Then, for each fixed $j\neq i$, we de-homogenise $g$ and $h$ with respect to the variable $X_j$, to obtain an expression for $f_{|U\cap U_j}$ as a rational function $r/s$ in $n$ variables, with denominator $s$ that does not vanish on $V\cap U_j$. The first step (homogenisation) is presented in a little more detail in the answer to question Q3 below.
A3: Suppose that $f \in \mathcal{O}_{U_i}(U)$. For each $x \in U$, there exists an open set $V$ with $x \in V \subseteq U$ such that under the identification $U_i \to \mathbb{A}^n(k)$, we have $f_{|V} = p/q$ with $p,q \in k[T_0,\ldots,T_{i-1},T_{i+1},\ldots,T_n]$ and $q(y)\neq 0$ for all $y\in V$. In accordance with this identification, we put $T_\ell = X_\ell/X_i$ for $0\leq \ell \leq n$, $\ell \neq i$, to obtain the expression $$ f_{|V}(X_0:\ldots:X_n) = \frac{p\left(\frac{X_0}{X_i},\ldots,\frac{X_{i-1}}{X_i},\frac{X_{i+1}}{X_i},\ldots,\frac{X_n}{X_i}\right)}{q\left(\frac{X_0}{X_i},\ldots,\frac{X_{i-1}}{X_i},\frac{X_{i+1}}{X_i},\ldots,\frac{X_n}{X_i}\right)}$$ of $f_{|V}$ as an element of $\operatorname{Map}(V,k)$. Now, we multiply the numerator and the denominator of the fraction on the right hand side by $X_i^d$, where $d = \max\{\deg p, \deg q\}$, in order to represent $f_{|V}$ as a fraction of the homogenous polynomials $$g = \Psi_i(p) = X_i^d\, p\left(\frac{X_0}{X_i},\ldots,\frac{X_n}{X_i}\right) \quad \text{and} \quad h = \Psi_i(q) = X_i^d\, q\left(\frac{X_0}{X_i},\ldots,\frac{X_n}{X_i}\right)$$ of degree $d$ in the variables $X_0,\ldots,X_n$ (we also have $h(y) \neq 0$ for all $y\in V$). The passage from $p/q \in k(T_0,\ldots,T_{i-1},T_{i+1},\ldots,T_n)$ to $g/h \in \mathcal{F} \subset k(X_0,\ldots,X_n)$ is the application of the inverse of the isomorphism given by (1.19.2).
A4: Let $g : U_i \to \mathbb{A}^n(k)$ denote the bijective map defined at the top of p. 28. The paragraph below it states that we define the space with functions $(U_i,\mathcal{O}_{U_i})$ by transport of structure from $(\mathbb{A}^n(k),\mathcal{O}_{\mathbb{A}^n(k)})$ via the map $g$. This means that a subset $U \subseteq U_i$ is open in $U_i$ if and only if $V := g(U)$ is open in $\mathbb{A}^n(k)$, and that a function $f \in \operatorname{Map}(U,k)$ belongs to $\mathcal{O}_{U_i}(U)$ if and only if it is of the form $p \circ g_{|U}$ for some $p \in \mathcal{O}_{\mathbb{A}^n(k)}(V)$, i.e. if and only if $f \circ {g^{-1}}_{|V} \in \mathcal{O}_{\mathbb{A}^n(k)}(V)$. Thus, $g$ is by definition an isomorphism of spaces of functions. Since Prop. 1.59 implies that $\mathcal{O}_{\mathbb{P}^n(k)}(U) = \mathcal{O}_{U_i}(U)$ as subsets of $\operatorname{Map}(U,k)$, there is nothing left to prove.