Is R/Z isomorphic to R/2Z?

It actually is correct! Taking $S^1$ and identifying antipodes gives $S^1$ again. This explains why $\mathbb{R}P^1$ is homeomorphic to $S^1$, and what you have is the group-theoretic version of this.

You have to be careful about how you are realizing $\mathbb{Z}_2$ as a subgroup of the circle. The sensible thing is as the multiplicative group $\{\pm 1\}$, and the cosets are then antipodal pairs, which gives my first paragraph above.

This is not a problem at all.

Consider the surjective group morphism $f: S^1 \to S^1$ $$f(z)=z^2$$ The kernel is $\{ 1 ; -1\} \cong \Bbb Z_2$ and by the first isomorphism theorem $$S^1 \cong S^1 / \{ 1 ; -1\}$$ which is exactly what you found.

Indeed, if you consider $S^1$ as the multiplicative group $\{ z\in\mathbb C : \lvert z\rvert = 1\}$, then $H=\{+1,-1\}$ is a subgroup isomorphic to $\mathbb Z/2\mathbb Z$ and you have an isomorphism \begin{align*} S^1/H &\longrightarrow S^1, \\ [z] &\longmapsto z^2, \end{align*} where $[z]=zH=\{z,-z\}$.