Let $\Sigma_{2}$ be a cyclic group of order 2 and let $\sigma$ be a nontrivial element of it. We can define an action of $\Sigma_{2}$ on $\mathbb{A}_{k}^{1}\backslash\{0\} = \mathrm{Spec}(k[x, y]/(xy-1))$ induced by $f(x, y)\mapsto f(-x, -y)$ for $f\in k[x, y]$. Note that this is just same as $a\mapsto -a$ for $0\neq a\in k$. Then the quotient $(\mathbb{A}^{1}_{k}\backslash\{0\})/\Sigma_{2}$ by the action can be identified with $\mathrm{Spec}((k[x, y]/(xy-1))^{\Sigma_{2}})$. Now we can show that there exists a surjective map $k[u, v, w]\to (k[x, y]/(xy-1))^{\Sigma_{2}}$ defined as $u\mapsto x^{2}, v\mapsto xy, w\mapsto y^{2}$, and so the quotient can be identified with a closed subset of $\mathbb{A}_{k}^{3}$.
However, I can't understand how this embedded subscheme $(\mathbb{A}_{k}^{1}\backslash \{0\})/\Sigma_{2}\hookrightarrow \mathbb{A}^{3}_{k}$ looks like. For example, if $k = \mathbb{C}$, then the quotient can be obtained by identifying $\mathbb{R}_{+}$ and $\mathbb{R}_{-}$ which are part of the boundary of the complex upper half plane and it may look like $\mathbb{A}^{1}_{\mathbb{C}}\backslash\{0\}$ again. However, I can't figure out what is the explicit map $(\mathbb{A}^{1}_{\mathbb{C}}\backslash \{0\})/\Sigma_{2}\to \mathbb{A}_{\mathbb{C}}^{3}$ and how image looks like in $\mathbb{A}_{\mathbb{C}}^{3}$. Actually, I want to know where the point $\overline{a}\in (\mathbb{A}^{1}_{\mathbb{C}}\backslash \{0\})/\Sigma_{2}$ maps to.