How should I visualize $S^n$ as the reduced suspension of $S^{n-1}$?

It is convenient to think of this in terms of the smash product $X \# Y$ of two spaces with base point. This is the space $X\times Y$ with $X \vee Y= X \times * \cup * \times Y$ shrunk to a point. Then the reduced suspension of $X$ is really $X \# S^1$. The general result is that

$$S^m \# S^n \cong S^ {m+n}.$$

One way of seeing this is to say that if $S^m$ has the cell structure $e^0 \cup e^m$, and similarly $S^n$, then $S^m \times S^n$ has the cell structure

$$e^0 \cup e^m \cup e^n \cup e^{m+n}, $$ and the first three cells determine $S^m \vee S^n$. So when you form the smash product $\ldots$

Here is a figure illustrating $S^1 \# S^1$ from Topology and Groupoids.

The reduced suspension $S S^{n-1}$ can be represented in $\mathbb{R}^{n+1}$ as the union of two cones on $S^{n-1} \subset \mathbb{R}^{n}\times\{0\},$ one with vertex at $(0,\ldots,0,1),$ and one with vertex at $(0,\ldots,0,-1).$

To get an explicit homeomorphism from this to $S^{n},$ you need not use stereographic projection; instead, just divide every point of $S S^{n-1}$ (each of which is a vector) by its norm.

EDIT: Since you're thinking about the reduced suspension, which I'll call $\Sigma S^{n-1},$ then your identification is probably the best one. You'll end up with a parabolic pencil of $(n-1)$-spheres about the north pole.

It is not hard to see that the unreduced suspension is homeomorphic to sphere, so you have to show that the result of collapsing a (standardly embedded) segment in a sphere is still a sphere.

Can you see what the result of collapsing one of the edges of a cube is?