Because the final map goes through 0, its image must be 0, but the final map is supposed to be the identity on $\mathbb{Z}$, which can't be the zero map.
Here is how it goes.
Let $B$, be a space nice enough to have a (simply connected) universal cover, say $B$ is connected, locally connected and semi-locally simply connected. Let $(X,x_0)\to (B,b_0)$ be its universal cover.
Take a loop $\gamma: (S^1,1)\to (B,b_0)$ then you can lift $\gamma$ to a path $\overline{\gamma}: I\to X$ that projects to $\gamma$. Now $\overline{\gamma}(1)$ is an element of $X_{b_0}$. You can use then the following theorem.
Let $(Y,y_0)\to (B,b_0)$ be a (path) onnected and locally path connected space over $B$ and $(X,x_0)\to (B,b_0)$ is a cover of $B$, then a lift of $(Y,y_0)\to (B,b_0)$ to $(Y,y_0)\to (X,x_0)$ exists iff the image of $\pi_1(Y,y_0)$ inside $\pi_1(B,b_0)$ is contained in the image of $\pi_1(X,x_0)$ inside $\pi_1(B,b_0)$
Use the previous theorem with $(Y,y_0)=(X,\overline{\gamma}(1) )$.
This tells you that there exists a covering map $X\to X$ sending $x_0$ to $\gamma(1)$.
It is easy to see that this map depends only on the homotopy class of $\gamma$ using the following result
Let $(X,x_0)$ be a cover of $(B,b_0)$ and $Y$ be a connected space over $B$. If two liftings of $Y\to B$ to $Y\to X$ coincide at some $y_0$ in $Y$, the they're equal.
This tells you that if $\overline{\gamma}(1)=\overline{\tau}(1)$ then the two morphisms $X\to X$ you get, coincide.
Moreover, using the inverse of $\gamma$, you see that the morphisms $X\to X$ you get are automorphisms.
This gives you a well defined map $\pi_1(B,b_0)\to \text{Aut}_B(X)$.
Using what I said before, it is easy to see that it is an isomorphism.
Best Answer
Hint: Since $S^n \cong D^n/\partial D^n$, a map $f': D^n \to \tilde S^n$ that maps $\partial D^n$ to a point is equivalent to a map $f'':S^n \to \tilde S^n$. And keep in mind that $p \circ f'=f$. (Click below for the full solution.)
Moral of the story: $\,$ In an ideal world, we would directly construct a lift $f'': S^n \to \tilde S^n$ of $\operatorname{id}_{S^n}: S^n \to S^n$, since this implies that $p_*$ is surjective. The problem is that we're not sure that such a lift exists because we don't know $\pi_1(S^n)$, so we have to pass through $D^n$.