It is easy to find continuous quotient maps that glue the boundary of the disk $D^n = \{x \in \mathbb{R}^n : \lVert x\rVert \le 1\}$ to one point to make $S^n = \{y \in \mathbb{R}^{n + 1} : \lVert y\rVert = 1\}$. (The norm here is the $l^2$ norm, which is smooth at all points except the origin) For example, we could take the map $x \mapsto \left(\frac{2\sqrt{\lVert x \rVert (1 - \lVert x\rVert)}}{\lVert x\rVert} x, 2\lVert x \rVert - 1\right)$. This is obviously continuous where $x \not= 0$, but is also continuous at $0$ because as $x$ goes to $0$, the image goes to $(0, -1)$. As $D^n$ is compact and $S^n$ is Hausdorff, this is a closed, and therefore, quotient map.
But this is not a smooth map since it contains the norm and square root, both of which are not smooth at $0$.
Is there a nice modification that gives a smooth quotient map? What about quotient maps from $\mathbb{R}^n$ to $S^n$?
Your idea of using stereographic projection (as indicated by the map you've provided) is a good one; we can fix things up by looking at a slight modification.
As you note, the norm map $x\mapsto \|x\|$ is not smooth at $0\in\mathbb{R}^n$. The square of this map, however, is smooth. Thus we can define a map $f\colon D^n\setminus\partial D^n\to\mathbb{R}^n$ by \begin{equation} f(x) = \dfrac{x}{1-\|x\|^2}, \end{equation} and this map will be smooth. We then have a stereographic projection map $g\colon \mathbb{R}^n\to\mathbb{R}^{n+1}$ given by \begin{equation} g(x) = \left(\dfrac{2x}{\|x\|^2+1},\dfrac{\|x\|^2-1}{\|x\|^2+1}\right), \end{equation} and this map is also smooth. The composition $g\circ f$ then gives us a map from $D^n\setminus\partial D^n$ to $\mathbb{R}^{n+1}$. Do you see how we could extend this composition to a map defined on all of $D^n$? Moreover, can you conclude that this gives a surjection $D^n\to S^n$, and that it plays nicely with the relevant topologies?