Here is the description:
Consider the boundary of the unit ball in $\mathbb{R}^n$, which is $S^{n-1}$. We have coordinate patches $\alpha:B^{n-1}(1)\to S^{n-1}, \textbf{u}\mapsto( \textbf{u},\sqrt{1-|| \textbf{u}||^2})$ and $\beta:B^{n-1}(1)\to S^{n-1}, \textbf{u}\mapsto( \textbf{u},-\sqrt{1-|| \textbf{u}||^2})$, where $B^{n-1}(1)$ is the open unit ball in $\mathbb{R}^{n-1}$.
On the one hand, direct computation shows that $D\alpha=\left[\begin{matrix}I_{n-1}\\D\sqrt{1-|| \textbf{u}||^2}\end{matrix}\right]$ and $D\beta=\left[\begin{matrix}I_{n-1}\\-D\sqrt{1-|| \textbf{u}||^2}\end{matrix}\right]$, hence the first $n-1$ rows of these two matrices are both $I_{n-1}$, hence they are the induced orientation of the natural orientation of the unit ball in $\mathbb{R}^n$.
On the other hand, we can define a orientation-reversing $r:\mathbb{R}^n\to\mathbb{R}^n,(x_1,\cdots,x_n)\mapsto(x_1,\cdots,-x_n)$ and obviously $\beta=\alpha\circ r$. Hence they belong to different orientations.
Therefore there is a contradiction within the discussion. But where did I make a mistake?
The coordinate patches, $\alpha $ and $\beta$, that you gave are the northern and southern hemispheres, in the case of $S^2$. (Setting them equal yields that the last coordinate is zero, so the intersection would be on the equator. But this is ruled out by the fact that charts are defined on open sets.) In general, the overlap is $\emptyset$. Therefore there is no issue of preserving orientation...
It takes $6$ of these types of charts to cover $S^2$. You should be able to generalize this to $S^n$ (it looks like $2(n+1)$ charts would do).
As I remarked, you could use stereographic projection (and only $2$ charts).