Consider in $\mathbb{R}^2$ the map $g(x,y) = (x+1,-y)$. It is easy to see that $G = \langle g\rangle$ acts proper and discontinously on $\mathbb{R}^2$, so $\pi : \mathbb{R}^2 \to \mathbb{R}^2/G$ is a covering map. I think that $\mathbb{R}^2/G$ is homeomorphic to $\mathbb{R}^2 \setminus \mathbb{D}^2$, where $\mathbb{D}^2$ is de unit disk. Is it correct?
Furthermore, we can parametrizes it with the cylinder $\mathbb{S}^1 \times [0,\infty)$ simply $[x,y] \mapsto (e^{2\pi i x},|y|)$.
Your answer is not correct because your "parameterization" is not one-to-one: given $(x,y) \in \mathbb R^2$, the elements $[x,y]$, $[x,-y] \in \mathbb R^2 / G$ are not equal, but their images in $\mathbb S^2 \times [0,\infty)$ under the map $[x,y] \mapsto (e^{2\pi i x},|y|)$ are equal.
To figure out what this quotient is, restrict the quotient map to a fundamental domain of the action of the group $\langle g \rangle$, namely to the infinite strip $[0,1] \times \mathbb R$. Its left edge $\{0\} \times \mathbb R$ is glued to its right edge $\{1\} \times \mathbb R$ by the identification $(0,y) \sim (1,-y)$. Does this quotient look familiar to you?