So first you want to know why an infinite string of spheres going off to infinity in both directions is the universal cover of $\Bbb RP^2\vee\Bbb RP^2$.
Let $\ast$ be the point of $\Bbb RP^2$ which comes from the points $N$ and $S$ of $S^2$ under the antipodal action, which are the north and south poles of $S^2$. Let $X$ be the space which is the disjoint union of two copies of $S^2$, $S^1\times\{0\}$ and $S^1\times\{0\}$, gluing their north poles with one another and their south poles with one another. Then we can consider the function $p:X\rightarrow \Bbb RP^2\vee\Bbb RP^2$ which on the first copy of $S^2$ of $X$ is the antipodal map onto the first copy of $\Bbb RP^2$ in $\Bbb RP^2\vee\Bbb RP^2$ and on the second copy of $S^2$ of $X$ onto the second copy of $\Bbb RP^2$ in $\Bbb RP^2\vee\Bbb RP^2$.
Then as the antipodal is a covering map, prove that $p$ is also a covering map; notice that the definition clearly holds for all points of $\Bbb RP^2\vee\Bbb RP^2$ different from $\ast$, so you have to see why it holds for $\ast$.
Now consider $\Bbb Z$ many copies of $S^2$, $S^2\times\{n\}, n\in\Bbb Z$, and let $X_\infty$ be the quotient obtained by gluing for each $n$ the north pole of $S^2\times\{n\}$ with the north pole of $S^2\times\{n+1\}$ and the south pole of $S^2\times\{n\}$ with the south pole of $S^2\times\{n-1\}$ if $n$ is even, and the north pole of $S^2\times\{n\}$ with the north pole of $S^2\times\{n-1\}$ and the south pole of $S^2\times\{n\}$ with the south pole of $S^2\times\{n+1\}$ if $n$ is odd.
Now consider the map $q:X\rightarrow X_\infty$ such that for each $n$, $q$ restricted to $S^2\times\{n\}$ is the identity map $S^2\times\{n\}\rightarrow S^2\times\{i\}$, where $i=0,1$ is such that $n\equiv i\mod 2$. Then it is easy to check $q$ is a covering map.
Thus $X_\infty$ is a covering space of $\Bbb RP^2\vee\Bbb RP^2$.
You also want to know why $X_\infty$ is simply connected. As $S^2$ is simply connected, it follows from Van-Kampen theorem and induction that any finite substring of $X_\infty$ is simply connected. Now any loop $\gamma$ in $X_\infty$ is contained in a finite substring of $X_\infty$, as $\gamma$ is compact and thus it is homotopic to the constant loop.
Finally, you ask why $S^2\vee S^2$ is not the universal cover of $\Bbb RP^2\vee\Bbb RP^2$. Well, the thing is that $S^2\vee S^2$ is not even a covering space of $\Bbb RP^2\vee\Bbb RP^2$. This is because $\Bbb RP^2\vee\Bbb RP^2$ is almost a surface except for the singularity $\ast$; which is a point having no open nhood homeomorphic to $\Bbb R^2$, so if $S^2\vee S^2$ were a covering space of $\Bbb RP^2\vee\Bbb RP^2$ it should have at leat two of this singularities, but $S^2\vee S^2$ has only one such singularity.