Find a basis on the $n$-sphere's tangent space $\mathcal{T}_s\mathbb{S}^n$ Let $\mathbb{S}^n$ be the $n$-sphere and $\mathcal{T}_s\mathbb{S}^n$ the tangent space at $\mathbf{s}\in\mathbb{S}^n$.
I am looking for a basis for the tangent $\mathcal{T}_s\mathbb{S}^n$, but I got confused.
Copying from Wikipedia, I understand that for a chart $\phi\colon\mathcal{U}\to\mathbb{R}^n$, where $\mathcal{U}$ is an open subset of $\mathbb{S}^n$, one can define an ordered basis as follows:

My questions are: a) can $\phi$ be the stereographic projection from $\mathbb{S}^n-\{N\}$ to $\mathbb{R}^n$ (where $N$ denotes the north pole of $\mathbb{S}^n$), and b) what can $f$ be in that case?
Thank you!
 A: It's easier by example. Let us think of $\mathbb S^2$, which is a subset of $\mathbb R^3$. Denote the points of $\mathbb R^3$ as $(X,Y,Z)$.
In the extrinsic viewpoint, the tangent space at $(0,0,1)\in \mathbb S^2$ is the set $\{(X, Y, 1)\ :\ X,Y\in\mathbb R\}$.
In the intrinsic viewpoint, you need a chart. You seem to like the stereographic projection. The equations of such projection from the South Pole $(0,0,-1)$ are
$$\tag{1}
x=\frac{X}{1+Z},\quad y=\frac{Y}{1+Z},$$
with inverse
$$\tag{2}
X=\frac{2x}{1+x^2+y^2},\quad Y=\frac{2y}{1+x^2+y^2},\quad Z=\frac{1-x^2-y^2}{1+x^2+y^2}.$$
The point $(X, Y, Z)=(0,0,1)$ corresponds to $x=0, y=0$. A basis of the tangent space at this point (and any other, except for the South Pole where the chart is undefined) is $\partial_x, \partial_y$.
For a $f\in C^\infty (\mathbb S^2)$ you can compute $\partial_x f|_{x=0, y=0}$ as follows. By definition, $f=f(X, Y, Z)$. So you plug in (2), yielding
$$
\frac{\partial}{\partial x} \left[ F\left( \frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}, \frac{1-x^2-y^2}{1+x^2+y^2}\right)\right]$$
and now you can compute and finally let $x=0, y=0$ in the result.
I hope this practical example clarified what is going on here.

Final notes.

*

*This answer is the continuation of my comment).


*The original question asked about the stereographic projection from the North Pole, whereas here it is from the South Pole. There is no essential difference.
