Application of Seifert-van Kampen Theorem I am trying to wrap my head around the following problem: I have three objects lined up horizontally, a $2$-sphere, a circle, and another $2$-sphere.  It is the wedge sum $S^2 \vee S^1 \vee S^2$.  I am trying to find the fundamental group of this space as well as the covering spaces. For the fundamental group, I believe that I can use van Kampen in the following manner: 
$$\begin{align*}
\pi_1(S^2 \vee S^1 \vee S^2) &= \pi_1(S^2) * \pi_1(S^1 \vee S^2) \\
\pi_1(S^2 \vee S^1 \vee S^2) &= 0 * \pi_1(S^1 \vee S^2) \\
\pi_1(S^2 \vee S^1 \vee S^2) &= 0 * (\pi_1(S^1) * \pi_1(S^2)) \\
\pi_1(S^2 \vee S^1 \vee S^2) &= 0 * \mathbb Z * 0 
\end{align*}$$
Does this make sense?  
I am still trying to work out how to find the covering spaces.
 A: You're correct about the computation of $\pi_1$, though depending on the level a bit more work may need to be shown (i.e., how does Seifert- van Kampen give the results you stated).  Also, $0*\mathbb{Z}*0$ is naturally isomorphic to $\mathbb{Z}$, and you may want to mention this.
As far as the covering space aspect, it may help to note that $S^2\vee S^1\vee S^2$ is homeomorphic to $S^1\vee S^2\vee S^2$ and you somehow need to "unravel" the $S^1$.  Since the universal cover of $S^1$ is $\mathbb{R}$, it should be no surprise that $\mathbb{R}$ enters the picture somehow when finding the universal covering space.
In fact, you might guess that the universal cover is $\mathbb{R}\vee S^2\vee S^2$, since this space is simply connected.  Unfortunately, this isn't correct.  The problem is that every $2\pi$ along the $\mathbb{R}$ piece should project to the wedge point which has an $S^2\vee S^2$ attached to it.  So, our next guess is that the universal cover is a copy of $\mathbb{R}$ with an $S^2\vee S^2$ attached to each point of the form $2\pi k$ for $k\in\mathbb{Z}$.  Now that you have the picture in mind, I'll leave it to you to try to prove this space is the universal cover.
More, in fact, is true:  If $X$ is simply connected, then the universal cover of $S^1\vee X$ is $\mathbb{R}$ with an $X$ wedged to each point of the form $2\pi k$.  Your proof in the $S^2\vee S^2$ case will likely automatically generalize to this statement.
A: The fundamental group of $S^2\vee S^1\vee S^2$ is $\mathbb{Z}$ as you stated ($S^2\vee S^2$ is simply connected, wedge with a circle gets you $\mathbb{Z}$). For each subgroup $n\mathbb{Z}$ of $\mathbb{Z}$ there is a covering space corresponding to that subgroup. These are "bracelets" with a sphere attached at regular intervals ($S^1$ with spheres attached), with $n\mathbb{Z}$ corresponding to the bracelet with $n$ beads. For $n=0$ the universal cover is an infinite bracelet ($\mathbb{R}$ with spheres attached).
This is assuming that the wedge points are distinct, else the "beads" on the bracelet come in pairs
