Is $\mathbb{R}$ a covering of $S^1 \vee S^1$? If so, doesn't this make it a universal covering? Hatcher gives $S^1 \vee S^1$ the universal cover  but doesn't the map $x \in (0,4\pi) \to \begin{cases} (\cos(x),\sin(x)) & 0<x\leq 2\pi\\ (2-\cos(x), \sin(x)) & 2\pi \leq x < 4\pi \end{cases}$ define a 1-fold covering and by extension $\mathbb{R}$ is a countable covering of $S^1\vee S^1$?
 A: Here's an answer to your question which compiles together the information in the comments. 
There does not exist any covering map $f : \mathbb R \mapsto S^1 \vee S^1$. 
For the proof, suppose that $f$ exists. It follows immediately from the definition of a covering map that for each $q \in S^1 \vee S^1$ and each $p \in f^{-1}(q)$ there exist open neighborhoods $V \subset S^1 \vee S^1$ of $q$ and $U \subset \mathbb R$ of $p$ such that $f$ restricts to a homeomorphism $f : U \to V$.
Now consider the case that $q$ is the "wedge point" of $S^1 \vee S^1$. Inside the open neighborhood $V$ of $q$ we may find another open neighborhood $V' \subset V \subset S^1 \vee S^1$ of $q$ such that $V'$ is connected and $V'-q$ has 4 connected components. Since $f^{-1} : V \to U$ is a homeomorphism, it follows that $U' = f^{-1}(V')$ is an open neighborhood $U' \subset U \subset \mathbb R$ of $p$ such that $U'$ is connected and $U'-p$ has 4 connected components. But $U'$ is an open interval, and $U'-p$ has only 2 connected components, a contradiction.

Now, there is still the issue of the map you propose; what is wrong with that map?
In order for your map $f$ to be a covering map, there must exist an arbitrarily small connected neighborhood $V \subset S^1 \times S^1$ of the wedge point $q$ such that for every connected component $U$ of $f^{-1}(V)$, the restricted map $f : U \to V$ is a homeomorphism. Your map does not satisfy that property. For instance, one may choose $V$ to be connected and so that $V-q$ has 4 connected components. But your domain $(0,4\pi)$ is an open subinterval, and every open connected subset of $(0,4\pi)$ is also an open subinterval and the complement of each point in that subinterval has 2 connected components.
