Let $U$ be the open rectangle $(0, \pi) \times (0,2 \pi) \subset \mathbb{R}^2 $ and let $X : U \rightarrow \mathbb{R}^3$ be the following map: $$X(\varphi , \theta)=(\sin \varphi \cos \theta , \sin \varphi \sin \theta , \cos \varphi ) $$ We can check whether $X$ preserves or reverses orientation by using the fact that it is the restriction of the 3-dimentional spherical coordinate parametrization $F : (0,1] \times U \rightarrow \bar {\mathbb{ B}}^3$ defined by $$F(\rho, \varphi, \theta)=( \rho \sin \varphi \cos \theta , \rho \sin \varphi \sin \theta , \rho \cos \varphi). $$ $F(1, \varphi, \theta) = X(\varphi , \theta) $ and the Jacobian determinant of $F$ is $\rho ^2 \sin \varphi$ , which is positive on $(0,1] \times U$. Thus $X$ is orientation-preserving.
This is a summary of Example 15.28 in Lee's Smooth manifolds(p. 388). The following lemmas might be used in this example.
An local diffeomorphism is orientation-preserving if and only if with respect to any oriented smooth charts, the Jacobian matrix has positive determinant.
Let $M$ be an oriented smooth n-manifold with boundary. Suppos $U \subset \mathbb{ R }^{n-1}$ is open, $a,b$ are real numbers with $a<b$, and $F :(a,b] \times U \rightarrow M$ is a smooth embedding that restricts to an embedding of $\{ b \} \times U$ into $ \partial M$. Then the parametrization $f: U \rightarrow \partial M$ given by $f(x)=F(b,x)$ is orientation-preserving for $\partial M$ if and only if $F$ is orientation-preserving for $M$.
In the Example, the Jacobian determinant to make sense, the standard coordinate chart should be suitable chart for $\mathbb{\bar B}^3$. But I think this is not.
$F$ is not even smooth.
A smooth chart for $\mathbb{\bar B}^3$ can be constructed by using stereographic projection as follows.(Problem 1-11. p. 31. Lee's Smooth Manifolds)
Stereographic projection $\sigma : \mathbb{S}^3 \setminus \{ (0,0,0,1) \}\rightarrow \mathbb{R}^3$
$$ \sigma(x,y,z,w)= \dfrac{(x,y,z)}{1-w}$$
Consider the map $\pi \circ \sigma^{-1}: \mathbb{R}^3 \rightarrow \mathbb{R}^3$, where $\pi$ is a projecton from $\mathbb{R}^4 $ to $\mathbb{R}^3$ that omits the third coordinate. Then $(\mathbb{\bar B}^3\setminus \{ (0,0,1)\} , (\pi \circ \sigma^{-1}|_{\mathbb{H}^3})^{-1})$ is a smooth chart.
For this chart the identity map $\mathbb{\bar B}^3\setminus \{ (0,0,1)\}$ (as a subset of $\mathbb{R}^3$) $\rightarrow \mathbb{\bar B}^3\setminus \{ (0,0,1)\}$ (as a manifold) has a coordinate representation of the form
$$(x,y,z) \rightarrow \dfrac{(x,y,\sqrt{1-x^2 -y^2-z^2})}{1-z} $$
For details, $$\pi \circ \sigma^{-1}|_{\mathbb{H}^3} (s,t,u) =\pi (\dfrac{(2s,2t,2u,s^2+t^2+u^2-1)}{s^2+t^2+u^2+1})=\dfrac{(2s,2t,s^2+t^2+u^2-1)}{s^2+t^2+u^2+1}$$ ,obtaining the inverse $$(\pi \circ \sigma^{-1}|_{\mathbb{H}^3})^{-1}(x,y,z)=\dfrac{(x,y,\sqrt{1-x^2 -y^2-z^2})}{1-z}. $$ This implies that if $x^2+y^2+z^2=1$ , this map doesn't smooth at these points.
Perhaps I have some misconceptions. I'd like to understand the Example thoroughly. Could I know where I mistake?
Your formulas are fine -- you're just a little mixed up about where these maps are going. The map $\pi\circ\sigma^{-1}$ takes $\mathbb R^3$ to $\mathbb R^3$, and takes the closed unit ball into the closed lower half-space $\{(x,y,z): z\le 0\}$, with boundary points going to the hyperplane $z=0$. Therefore $\varphi = \pi\circ\sigma^{-1}|_{\overline{\mathbb B}{}^3}$ is a candidate for a boundary chart for the closed ball, at least wherever it's a local diffeomorphism. (OK, well, technically a boundary chart should map into the upper half-space, but that's easily fixed.) You can check that this applies to any point where $u\ne 0$.
The relevant points in the image of $\varphi$ are those for which $z=0$ and $x^2+y^2<1$, which are images of boundary points of $\overline {\mathbb B}{}^3$ under this boundary chart. At all such points, $\varphi^{-1}$ is smooth.