Group Action of $\mathbb{Z}$ on $\mathbb{R}^2\setminus \{0\}$ and covering space if the group $G=\mathbb{Z}$ acts on $Y=\mathbb{R}^2\setminus \{0\}$ using by the action $m\cdot (x,y)=(2^m x, 2^m y)$ then I was able to show that
$Y \to Y/ \mathbb{Z}$ is a covering  map.
Can I conclude that $\pi_1(Y/\mathbb{Z})=G=\mathbb{Z}?$
I was able to show that the map is covering map using the fact that a properly discontinuous map gives us a covering map.
 A: You need to state what results are letting you conclude that $\pi_{1}(Y/\Bbb{Z})\cong \Bbb{Z}$ . Ofcourse it is not at all true that $\pi_{1}(Y/G)\cong \Bbb{Z}$. Technically you need to produce an explicit isomorphism between the two groups. Ofcourse if you already know results then fine but still try to state them clearly. In doing so many wrong things will get cleared out.
For example the fact you are using is incorrect. You have $\frac{\pi_{1}(Y/G,y)}{p_{*}(\pi_{1}(Y,y))}\cong G$ and not what you state.See Hatcher Proposition $1.40$ pg $72$ for a proof .What you only have is that the quotient map is a covering map from Y to the orbit space. Also try not to write $\pi_{1}(X,x)$ is equal to $\Bbb{Z}$ . Write it as $\pi_{1}(X,x)\cong \Bbb{Z}$. That is because you have an isomorphism and not necessarily an equality of fundamental groups. Sometimes we just abuse notation and write equal to.
Solution:
We will solve the problem by proving that $Y/G$ is homeomorphic to the Torus $S^{1}\times S^{1}$.
$\Bbb{R}^{2}\setminus 0\cong (0,\infty)\times [0,2\pi]/0\sim 2\pi = (0,\infty)\times S^{1}$ by using a variable change to polar coordinates.
Then the action of $\Bbb{Z}$ becomes $m\cdot(r,\theta)=(2^{m}r,\theta)$ .
Now $\log_{2}(x):(0,\infty)\to\Bbb{R}$ is a homeomorphism.
Under this homeomorphism the action becomes $m\cdot(r,\theta)=(\log_{2}(r\cdot 2^{m}),\theta)=(m+\log_{2}(r),\theta)$.
That is the action of $\Bbb{Z}$ on $\Bbb{R}$ just becomes $m\cdot x = x+m$ .
Using this, the action of $\Bbb{Z}$ just factors to $\Bbb{R}^{2}\setminus\{0\}/\Bbb{Z}\cong (0,\infty)/\Bbb{Z}\times S^{1}\cong \Bbb{R}/\Bbb{Z}\times S^{1}$ .
Now it's an easy excercise to show that $\Bbb{R}/\Bbb{Z}$ such that $x\sim y\iff x-y\in\Bbb{Z}$ is just homeomorphic to $S^{1}$.
Define $e:\Bbb{R}\to S^{1}$ by $e(x)=e^{i2\pi x}$ . Then $x\sim y\iff x-y\in\Bbb{Z}\iff e(x)=e(y)$ . Then this map descends to a unique homeomorphism from the quotient to prove that $\Bbb{R}/\Bbb{Z}\cong S^{1}$
So $Y/G\cong S^{1}\times S^{1}$ which is a torus.
So the fundamental group is isomorphic to $\Bbb{Z}\times \Bbb{Z}$.
