The non-existence of continuous functions on $\mathbb{R}$ which attain each value in their range exactly twice has been raised numerous times, yet I cannot find (either here or elsewhere) analogous theorems for $\mathbb{R}^n$ or other metric/topological spaces.
All I can observe is that on $\mathbb{C}-\{0\}$, a counterexample would be $f(z)=z^2$, but I do not know what to say if the origin is added back.
Assume that you have a map $\mathbb{R}^n\to Y$ for some topological space $Y$, such that $|f^{-1}(y)|=2$ for all $y\in Y$.
I'll consider the particular case where you make the additional assumption:
The map $f$ is "nice" in the sense that for all $y\in Y$, the two elements of $f^{-1}(y)$ are distant by more than some absolute constant $\varepsilon> 0$.
Then, consider $x\in \mathbb{R}^n$ and the restriction of $f$ to the open ball $B(x,\frac{\varepsilon}{2})$. Since bounded closed subsets of $\mathbb{R}^n$ are compact, this restriction is a closed map. It is also continuous and bijective onto its range, therefore it is a homeomorphism onto its range. It follows that $f$ is a covering map.
Now since $\mathbb{R}^n$ is simply connected, $f$ is a universal cover of $Y$, and since the fiber has two elements, it follows that $$\pi_1(Y)\simeq \mathbb{Z}_2$$
So there you are, under a pretty reasonable restriction on $f$, you see that the target space needs to have a very particular topology - it can't be $\mathbb{R}^n$ itself for instance.