Showing that $(\mathbb{C}^{*}/\mathbb{R}^{*},\cdot)\cong (U/U_2,\cdot)$ I want to show that $(\mathbb{C}^{*}/\mathbb{R}^{*},\cdot)\cong (U/U_2,\cdot)$, where $U$ is the unit circle and $U_2=\{-1,1\}$.
I considered the function $f:\mathbb{C}^{*}\to (U/U_2,\cdot)$, $f(z)=\hat{\frac{z}{|z|}}$.
Then $f$ is well defined and it is a surjective homomorphism.
Let $z\in \ker(f)$. Then $f(z)=\hat{1}\iff \hat{\frac{z}{|z|}}=\hat{1}\iff\frac{z}{|z|}\in U_2\iff z\in \mathbb{R}^{*}$. So $\ker(f)=\mathbb{R}^{*}$ and now we are done by the fundamental isomorphism theorem.
I think that this proof should work, but I want to ask a further question. If I considered $g:\mathbb{C}^{*}\to U$, $g(z)=\frac{z}{|z|}$, I think that I would have got in the same manner that  $(\mathbb{C}^{*}/\mathbb{R}^{*},\cdot)\cong (U,\cdot)$. This together with the other isomorphism implies that $(U/U_2,\cdot) \cong (U,\cdot)$.
So, could I say that factoring by $U_2$ is basically pointless?
 A: Yes, modding out by $U_2$ is "trivial".
Note that if $G \leq S^1$is a finite subgroup, then it consists of roots of unity. One can actually show that $G = G_n$ consists of all the $n$-th roots of unity for some $n \geq 1$.
But $z \mapsto z^n$ is a surjective endomorphism of the sphere, and thus $S^1/G_n \simeq S^1$.
Edit: your second map has kernel $z \in \Bbb C^\times$ such that $z = |z|$. That's not $\Bbb R^\times$ but rather $\Bbb R_{> 0}$.
Here's another way to see your first construction: you first map gives an iso
$$
\tau \colon \Bbb C^\times/\Bbb R_{>0} \to S^1.
$$
The preimage of $G_2$ are precisely the classes $[z]$ for which $z/|z| = \pm 1$, so $z = \pm|z|$ i.e. $z \in \mathbb{R}^\times$. Inside the quotient $\Bbb C^\times /\mathbb{R}_{>0}$ that's just $[-1]$ and $[1]$, but we can also think of it as $\Bbb R^\times /\Bbb R_{>0}$, hence
since $\tau(\Bbb R^\times / \Bbb R_{>0}) = G_2$ we get
$$
\Bbb C^\times / \Bbb R^\times  \simeq (\Bbb C^\times/ \Bbb R_{>0}) / (\Bbb R^\times / \Bbb R_{>0}) \simeq S^1/G_2. 
$$
This is still $S^1$, though. An explicit map $\Bbb C^\times / \Bbb R^\times \to S^1$ would be induced by $z \mapsto z^2/|z|^2$.
