We first present a pedantic construction, then we introduce a simpler way of making the same construction.
Constructing $\mathbb{P}^n$ by gluing.
Classical Construction
Before proceeding to the scheme case, let us look at the the classical $$\mathbb{P}^n(k) = \{[X_0: \ldots: X_n] \mid X_i \text{ not all $0$} \}/ \sim.$$
We see that $\mathbb{P}^n$ is covered by pieces $U_i = \{ X_i \neq 0\}$. We can identify each $U_i$ with affine space $k^n$ (i.e. put affine coordinates on $U_i$) via the map $\phi_i: U_i \to k^n$ given by
$$[X_0: \ldots: X_n] \mapsto (X_{0}/X_i, \ldots, \widehat{X_i/X_i}, \ldots, X_n/X_i).$$
If $x \in U_i \cap U_j$, what is the relationship between the coordinate $\phi_i(x)$ and $\phi_j(x)$?
Suppose $(x_{0/i}, \ldots, \widehat{x_{i/i}}, \ldots, x_{n/i})$ denotes the coordinate of $\phi_i(x)$. Then we must have $\phi_j(x) = (\frac{x_{0/i}}{x_{j/i}}, \ldots, \widehat{x_{i/i}}, \ldots, \frac{x_{n/i}}{x_{j/i}}).$
In other words, we have a map between coordinate rings of $U_i$ and $U_j$, viewed as affine varieties under identification $\phi_i$ and $\phi_j$ given by
$$A(U_j) = k[x_{0/j}, \ldots, \widehat{x_{j/j}}, \ldots, x_{n/j}] \to A(U_i) = k[x_{0/i}, \ldots, \widehat{x_{i/i}}, \ldots, x_{n/i}] $$
$$x_{0/j} \mapsto \frac{x_{0/i}}{x_{j/i}}$$ and so on.
Gluing Schemes
Motivated by the above discussion, for $i = 0$ to $n$, we let $X_i = \mathbb{A}^n = \text{Spec} k[x_{0/i}, \ldots, \widehat{x_{i/i}}, \ldots, x_{n/i}] = \text{Spec} \left(k[x_{0/i}, \ldots, x_{n/i}] /(x_{i/i} - 1)\right).$
Let $$X_{ij} = \text{Spec} \left(k\left[\frac{x_{0/i}}{x_{j/i}}, \ldots, \frac{x_{n/i}}{x_{j/i}} \right]/(x_{i/i}- 1)\right) = \text{Spec}\left(k\left[x_{0/i}, \ldots, x_{n/i}, \frac{1}{x_{j/i}}\right]/(x_{i/i} - 1) \right) =: \text{Spec} A_{ij}$$
be an open subset of $X_i$.
We have the following identification $\varphi_{ij}: X_{ij} \to X_{ji}$ corresponding to the ring homomorphism $A_{ji} \to A_{ij}$ given by
$$ x_{0/j} \mapsto \frac{x_{0/i}}{x_{j/i}}$$
and so on.
Note that this map is well defined since $x_{j/j} - 1$ is mapped to $\frac{x_{j/i}}{x_{j/i}} - 1 = 0$.
It is easy to see that $\varphi_{ij}$ and $X_{ij}$ satisfies the cocycle condition. This is because
$X_{ij} \cap X_{ik}$ is affine and equal to $\text{Spec}(A_{ijk})$ where
$$A_{ijk} = k\left[x_{0/i}, \ldots, x_{n/i}, \frac{1}{x_{j/i}}, \frac{1}{x_{k/i}}\right]/(x_{i/i} - 1).$$
As before, the map $\varphi_{ij}|_{X_{ijk}}: X_{ijk} \to X_{jik}$ (restriction of $\varphi_{ij}$) corresponds to the map $A_{jik} \to A_{ijk}$ (extension of the previous map) given by
$$x_{0/i} \mapsto \frac{x_{0/i}}{x_{j/i}}.$$
In particular $\frac{1}{x_{j/i}} \mapsto 1$ and $\frac{1}{x_{k/i}} \mapsto \frac{x_{k_i}}{x_{j/i}}.$
On the other hand the map $\varphi_{kj} \circ \varphi_{ik}|_{X_{ijk}}$ corresponds to the map
$A_{ijk} \to A_{kji} \to A_{jik}$ given by
$$x_{0/i} \mapsto \frac{x_{0/i}}{x_{k/i}} \mapsto \frac{\frac{x_{0/i}}{x_{k/i}}}{\frac{x_{j/i}}{x_{k_i}}} = \frac{x_{0/i}}{x_{j/i}}.$$
Thus we can glue $X_i$ along $X_{ij}$ via $\varphi_{ij}$ to get a unique scheme $\mathbb{P}^n$.
In the interpretation of Bosch, the second equalitz concerning A_i is not true. The first one is the homogeneous localization at x_i but the second one is the unconditional localization.
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