For your first question: using the stereographic projection you can project an $n$-dimensional sphere onto (what is called) a hyperplane, the $(n-1)$-dimensional object of your question.
For your second question, let's start defining (in one of the many possible ways) the stereographic projection:
Consider the unit sphere in the $3$-dimensional space $\mathbb{R}^3$ and let $N = (0, 0, 1)$ be the north pole and $S=\mathbb{S}^2\setminus\{N\}$ be the rest of the sphere. The plane $z = 0$ runs through the center of the sphere and the equator is the intersection of the sphere with this plane. For any point $p$ on $S$ there is a unique line through $N$ and $p$, and this line intersects the plane $z = 0$ in exactly one point $p^\prime$. We define the stereographic projection of $p$ to be this point $p^\prime$ in the plane.
As explained in the video of your question, the stereographic projection has nice properties like the conformality, meaning that it preserves the angles, despite it does not preserve area. In such a projection, circles on the sphere that do not pass through the point of projection are projected to circles on the plane and circles on the sphere that do pass through the point of projection are projected to straight lines on the plane.
I said this is one of the many ways we can define it, and you can check here why and how to define others. Using this idea, I will try to answer your second question:
As you said in the $3$-dimensional space the notion of north pole is basically the point on the sphere with maximum $z$-coordinate for a given coordinate system and this can be extended to any dimension by just considering the point $(0,0,0\dots,0,1)$ as the distinguished north pole.
But you don't really need this point. In general, one can define a stereographic projection from any point $w\in\mathbb{S}^n$ on the $n$-sphere onto any hyperplane $\Pi\in\mathbb{R}^{n+1}$ such that
- $\Pi$ is perpendicular to the diameter through $w$
- $\Pi$ does not contain $w$.
As long as $\Pi$ meets these conditions, then for any point $p$ other than $w$ (our distinguished point now) the line through $p$ and $w$ meets $\Pi$ in exactly one point $p^\prime$, which is defined to be the stereographic projection of $p$ onto $\Pi$. I use the terminology stereographic in the previous sentence because this projection has the same essential properties as the one in the $3$-dimensional case from the north pole $N$: they are smooth bijections with smooth inverse defined everywhere except at the projection point and they are conformal and not area-preserving.
Your formulas are correct (as shown in Riemannian Manifolds:
An Introduction to Curvature by Lee 1997, Formula 3.9).
Remember that to prove that $\sigma^{-1}(\sigma(x))=x$ for $x \in S^n_R$ you need the fact that $R^2 = ||x_{1:n}||^2 + x_{n+1}^2$, where $R=1/\sqrt{K}$ in your formulas.
Best Answer
In Cartesian coordinates it's $(x_1,\dots,x_{n+1})\to (\frac{Rx_1}{R-x_{n+1}},\dots,\frac {Rx_n}{R-x_{n+1}},0)$.
See here.