The unit sphere $n$ dimensional is the set $$\mathbb{S}^n=\bigg\{(x_1,x_2,\dots, x_{n+1})\in\mathbb{R}^{n+1}\;|\;\big(x_1^2+x_2^2+\cdots+x_{n+1}^2\big)^{1/2}=1\bigg\}.$$ For all $i=1,\dots, n+1$ denote with $U_i^+$ the subset of $\mathbb{R}^{n+1}$ so defined $$U_i^+=\bigg\{\big(x_1,x_2,\dots, x_{n+1}\big)\in\mathbb{R}^{n+1}\;|\; x_i>0\big)\bigg\}.$$
We consider the continous function $f\colon\mathbb{B}^n\to \mathbb{R}$ so defined $u\mapsto\sqrt{1-\lvert u\rvert^2}$. Then for all $i=1,\dots, n+1$, $U_i^+\cap \mathbb{S}^n$ is the graph of the function $$x_i=f(x_1, x_2,\dots, x_{i-1},x_{i+i},\dots, x_{n+1}).$$
Let's prove explicitly what has just been said.
In symbols, pointing with $\hat{x}=(x_1,\dots, x_{i-1}, x_{i+1},\dots, x_{n+1})$, for all $i=1,\dots, n+1$ we have $$U_i^+\cap \mathbb{S}^n=\bigg\{x=(\hat{x}, x_i)\in\mathbb{R}^n\times\mathbb{R}\;|\; \hat{x}\in\mathbb{B}^n,x_i=f(\hat{x})\bigg\}.$$ Indeed, if $x\in U_i^+\cap \mathbb{S}^n$, then $x_i>0$ and $\lvert x \rvert =1$ On $U_i^+$ can resolve the equation $\lvert x \rvert=1$ respect to $x_i$ then we have $$x_i=\sqrt{1-(x_1)^2- \cdots -(x_{i-1})^2-(x_{i+1})^2- \cdots - (x_{n+1})^2}.$$ If we consider the vector $\hat{x}=(x_1, x_2,\dots, x_{i+1}, x_{i+1}, \dots, x_{n+1})$ we have $\lvert \hat{x} \rvert <1$, then $\hat{x}\in\mathbb{B}^n$, therefore $x=(\hat{x}, x_i)\in U_i^+\cap\mathbb{S}^n$. Vice versa, if $x=(\hat{x}, x_i)\in U_i^+\cap \mathbb{S}^n$, then $\hat{x}\in\mathbb{B}^n$ and $x_i=\sqrt{1-\lvert \hat{x} \rvert}>0$, moreover $\lvert x \rvert=1$, therefore $x\in\mathbb{S}^n$
Question. Is my reasoning formally correct?
I think your book uses strange notation, and is possibly wrong. The graph $\Gamma_f$ of $f: A \longrightarrow B$ is defined as the subset of $A \times B$ given by $\Gamma_f = \{(a,b) \in A\times B\mid b = f(a)\}$.
The equation $x_i = f(x_1,x_2,\dots,\hat{x_i},\dots,x_n, x_{n+1})$ is not a function and hence does not have a graph, but it implicitly defines a subset $S_i$ of $\Bbb R^{n+1}$ via
$$S_i = \left\{(x_1,\dots,x_{n+1})\in\Bbb R^{n+1}\,\middle|\, x_i = f(x_1,x_2,\dots,\hat{x_i},\dots,x_n, x_{n+1})\right\}.$$
Moreover, as I pointed out in the comments, the book appears to be wrong. There are points in $S_i$ whose $i$-th coordinates is $0$.
Indeed, if $H_i\subset R^{n+1}$ is the hyperplane $\left\{(x_1,\dots,x_{n+1})\in\Bbb R^{n+1}\,\middle|\, x_i =0\right\}$, then any $p\in H_i \cap \Bbb S^n$ satisfies the equation $x_i = f(x_1,x_2,\dots,\hat{x_i},\dots,x_n, x_{n+1})$ and hence lies in $S_i$.
No such $p$ however lies in $U_i^+$, so the claim made by the book cannot be right.