This is slightly too long for a comment, therefore I'm posting it here:
In complete generality: Let $S$ be a subset of a group $G$. Then you're familiar with the subgroup $\langle S \rangle$ generated by $S$. This is the smallest subgroup of $G$ containing $S$. Similarly, there is $N = \langle \langle S \rangle \rangle$, the smallest normal subgroup generated by $S$ (sometimes also called normal or conjugate closure). While $\langle S \rangle$ consists of precisely the words of the form $s_{i_1}^{\pm 1} \cdots s_{i_{n}}^{\pm 1}$ with $s_{i_{j}} \in S$, the smallest normal subgroup consists of the words of the form $g_{1}s_{i_1}^{\pm 1}g_{1}^{-1} \cdots g_{n}s_{i_{n}}^{\pm 1}g_{n}^{-1}$ with $g_{j} \in G$.
For example, if you're writing $G = \langle S | R \rangle$, i.e. $G$ is given in terms of a presentation with generators $S$ and relations $R$, then you actually mean $G \cong F(S) / \langle\langle R \rangle \rangle$, or in words $G$ is the quotient of the free group on $S$ modulo the normal soubgroup generated by the relations. The normal subgroup generated by a set is tremendously hard to determine (that's why the Word problem is so difficult, i.e. not solvable in general).
Now you should be able to understand what $G \ast_{A} H$ is: it's $G \ast H/ \langle\langle R \rangle \rangle$, with $R = \{\varphi(a)\psi(a)^{-1}\,:\,a \in A\}$, where $\varphi:A \to G$ and $\psi: A \to H$ are the given homomorphisms.
In your concrete situation, the situation is quite silly: As $G \ast H = \{1\}$, we must have that $N = \{1\}$.
If you want to learn about free products with amalgamation in general, the standard reference is J.-P. Serre's Arbres, amalgames et $SL_{2}$ (translated as Trees). As for Seifert-van Kampen, I think Allen Hatcher has a pretty lucid and detailed explanation on pages 40ff of his algebraic topology book, available on his home page.
I recall enjoying the van Kampen exercises in grad school, so I will give this one a try. The topologists can hopefully add more helpful answers.
As the first example consider the sphere with a single handle, i.e. the torus. You hopefully already know the answer, so let's see how van Kampen does it. We are to split the torus into two parts, $U$ and $V$ with a path-connected intersection, such that we know the fundamental group of $U,V$ and $U\cap V$. Let $V$ be just a small open disk on the surface of the torus. Let $K$ be an even smaller closed disk inside $V$, and let $U$ be the complement of $K$ on the entire torus, so $U\cap V = V\setminus K$ is an open annulus around the perimeter of $V$.
$V$ is contractible and has a trivial fundamental group.
$U\cap V$ contracts to a circle, and has the infinite cyclic group as the fundamental group. A generator of this group is the loop $g$ going once around the circle.
$U$ is essentially the torus with a hole in it made by the removal of the patch $K$. By making that hole bigger and bigger, we eventually see that $U$ contracts to a figure eight 8. If we look at the torus as a bicycle tube, then $U$ is the tube with the valve and its surroundings removed, and it contracts to the union of a big circle $x$ (the points that would touch the ground, if you rotate the wheel 360 degrees) and one small circle $y$ around the tube. These two circles intersect at a single point. So the fundamental group of $U$ is the free group on two generators $x$ and $y$.
What does van Kampen tell us about the fundamental group of the union $U\cup V$? Basically we get the free product of the fundamental groups of $U$ and $V$, but we need to do a bunch of identifications by introducing relations that equate the images (under the induced by the inclusion map) of the elements of $\pi_1(U\cap V)$ on either side.
Here $\pi_1(U\cap V)=\langle g\rangle$, so we only need to check, what kind of a relation we get by equating the image of $g$ in $\pi_1(U)$ and $\pi_1(V)$. In $\pi_1(V)$ the image of $g$ is, of course, trivial, because the loop trivially contracts to a point, once we allow it to pass into $K$. The fun part is to observe that when we "expand the hole $K$ to the complement of the figure 8", the loop $g$ becomes the commutator $xyx^{-1}y^{-1}$ (alter the order of the factors depending on the choices of the orientations that you made).
You may need to draw a picture to see this happening. As a substitute I would suggest that if we form the torus by glueing together the opposite sides of a rectangle in the usual way, then $K$ is a hole in the middle, $g$ is a loop around it, the figure 8 is the border of the rectangle with the obvious identifications, and "expanding the hole" results in pressing the loop to go once around the perimeter of the square.
Anyway, van Kampen tells us now to identify $xyx^{-1}y^{-1}$ with the trivial element turning the free group on two generators into a free abelian group on two generators. The powers of $g$ don't introduce any new relations, so we are done.
Best Answer
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.