Your space is homotopy equivalent to the sphere $S^2$ with a line segment (in blue in the picture) joining $p$ and $q$:
By continuously deforming the line segment, you can fuse the endpoints and you end up with $S^2 \vee S^1$, the wedge of a sphere and a circle at a point (I've drawn the circle outside the sphere, it doesn't change anything):
Since the sphere $S^2$ is simply connected, the universal cover of this space looks like a "necklace". It's an infinite (in both directions) number of spheres, and each sphere is linked to the next by a line segment:
Each sphere is mapped homeomorphically onto the $S^2$ component of $S^2 \vee S^1$, and each segment is projected onto the circle component by identifying endpoint.
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
Not in general. This fails to be true for odd dimensional spheres or when dealing with non-Deck maps. But ultimately it is true in this case.
No, Deck transformations are not defined up to homotopy. These are concrete, special homeomorphisms together with function composition. You are right that $g\circ f\simeq id$ but this is not enough for us. In general, Deck transformations can be homotopic but not equal (and in such case they will correspond to different elements of $\pi_1(X)$). But of course not in our case.
You are very close. You just need a purely algebraic argument. In current situation $\mathrm{deg}:G\to\{-1,1\}$ is a group homomorphism. And so the condition "$\mathrm{deg}(f)=-1$ when $f\neq id$" means that the kernel of $\mathrm{deg}$ is trivial. And therefore $\mathrm{deg}$ is injective, which finally means that (up to isomorphism) $G$ has to be a subgroup of $\{-1,1\}$.