Let $p \in \mathbb{S}^{n}$, then the stereogaphic projection is a diffeomorpshim $h:\mathbb{S}^{n} \setminus \{p\} \to \mathbb{R}^{n-1}$.
Suppose that $p$ is the 'north pole' ($p = (0,0,..,1)$), then $h$ is easily computable with the formula given by $[h(x)]_{i} = \frac{[x]_{i}}{1 - [x]_{n}}$, for $x \in \mathbb{S}^{n}$ and $1 \leq i \leq n-1$.
However, if $p$ is some arbitrary point, then $h$ seems somewhat more elusive. Is there a general way to express $h$ in such a case?
Obviously, one could rotate the sphere such that $p$ is at the pole. But, that doesn't seem very efficent.
Stereographic projection when the "North/South Pole" is not given by $(0,...,\pm 1)$? Gives an answer, but the image of the function is embedded in the hyperplane orthgonal to $p$ and not in $\mathbb{R}^{n-1}$ itself.
You could do a stereographic projection from the north pole, followed by a Möbius transformation. A Möbius transformation is determined by the images of three points. You'd want to map the image of $p$ to the point at infinity, the image of the antipode of $p$ to the origin, and any point on the equator polar to $p$ to the unit circle. All of this has the same effect as performing the rotation in space, but depending on your background you might consider it easier.
If you don't care for changes due to translations or scaling, then it would be enough to map the image of $p$ whcih I'll call $p'$ to the point at infinity, e.g. using $z\mapsto\frac1{z-p'}$.