The $n$-disk $D^n$ quotiented by its boundary $S^{n-1}$ gives $S^n$

None of the answer give any map explicitely so I'll write one here. Note that $S^{n−1}\times I→D^n,(u,t)\mapsto tu$ (this is scalar product) is a quotient map and thus one can define (after verifying some things) the continuous map $D^n\to S^n,tu\mapsto (\sin⁡(tπ)u,\cos⁡(tπ))$ which can be verified to work.

Yes, you only need to find a suitable map between $D^n$ and $S^n$, and to do that, your visualisation should help you a lot. For this kind of problem, getting a visualisation is most of the work, and you did that already. The rest is just describing what you see.

Let's pick $n=2$. You said that if you push "down" the center of $D^2$ and "close" the sack, you would get $S^2$. This should tell you that you are looking at a map $f : D^2 \rightarrow S^2$ such that the $f(0,0) = (0,0,-1)$ and is $f(x,y) = (0,0,+1)$ for $(x,y) \in S^1$. Then you need to pick a way to choose for example the $z$ coordinate for intermediate points in a continuous way. Any continuous function should work.

Do you visualise your sack with a round shape all the time ? What should be the inverse image of the equator ? A circle of center $(0,0)$ in $D^1$ ? Then your map should be rotationnaly invariant around $(0,0)$ and you can settle to search for $f$ only on a diameter of $D^2$. What is the image of a diameter in your visualisation ? A circle of $S^2$ ? So $f(x,y)$ should be $(kx,ky,z)$ for some functions $z$ and $k$ ? etc.

Since $(D^n\setminus S^{n-1}) \cong \mathbb{R}^n \cong (S^n-\{p\})$, we have a homeomorphism $k:D^n\setminus S^{n-1} \rightarrow S^n-\{p\}$. Then we can define $f(x)=k(x)$ for $x \in D^n\setminus S^{n-1}$ and $f(x)=p$ for $x \in S^{n-1}$.

The continuity of $f$ can be shown as follows. Let $U$ be an open proper subset of $S^n$ containing $p$. Then $U \cong \mathbb{R}^n \cong (D^n\setminus S^{n-1})$. Simultaneously removing a closed subset of each of the spaces yields $U\setminus\{p\} \cong \mathbb{R}^n\setminus D^n \cong (D^n\setminus S^{n-1})\setminus C$, where $C\subset D^n\setminus S^{n-1}$ is closed. So we can say $f^{-1}(U\setminus \{ p \} ) = (D^n\setminus S^{n-1})\setminus C$, since on the domain $U\setminus \{ p \}$, $f$ is a homeomorphism. Noting that any open neighborhood of $S^{n-1}$ must be of the form $D^n\setminus C$, it is easy to see that $f^{-1}(U)$ is of such form and open. Then $f$ is the desired map.