I'm looking for ways to find the fundamental group of $\mathbb{T}^2\vee\mathbb{S}^1$.
I already know how to find $G:=\pi_1(\mathbb{T}^2\vee\mathbb{S}^1)$ using van Kampen's theorem with open sets $A=\mathbb{T}^2\cup(\text{small open of $\mathbb{S}^1$})$ and $B=\mathbb{S}^1\cup(\text{small open of }\mathbb{T}^2)$ so that:
$$G=\pi_1(\mathbb{T}^2)*\pi_1(\mathbb{S}^1)\simeq (\mathbb{Z}\times\mathbb{Z})*\mathbb{Z}$$
Now I'm trying to calculate it by covering spaces. Using the fact that the universal covers of $\mathbb{T}^2,\mathbb{S}^1$ are $\mathbb{R}^2,\mathbb{R}$ respectively, I believe the fundamental cover $E$ of $\mathbb{T}^2\vee\mathbb{S}^1$ looks like this:
Considering $E\subset\mathbb{R}^3$, then each $(a,b,c)\in\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ gives a homeomorphism $E\to E$ by $(x,y,z)\mapsto (x+a,y+b,z+c)$. So the covering map can be seen as $E\to E/(\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z})$, so that $G=\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\neq (\mathbb{Z}\times\mathbb{Z})*\mathbb{Z}$.
Obviously I've made a mistake, but I still can't see where.
EDIT: After reading Eric Wofsey's answer, I've made a second try:
(The orange dots in the domain are the integer points in $\mathbb{R}^2$ and are sent to the orange dot in the image)
Best Answer
The space $E$ you have drawn does not cover $T^2\vee S^1$. Notice in particular that every integer lattice point in each of the planes should map to the basepoint, but locally near those lattice points (except for the origin) $E$ looks like $\mathbb{R}^2$, rather than $\mathbb{R}^2\vee\mathbb{R}^1$. To get the correct universal cover you would need to have more lines branching off of each of the lattice points, and then more planes along each of those lines, and more lines along the lattice points in those planes, and so on. This is fairly complicated to write out in full detail but if you do so you will find that the deck transformation group is indeed $(\mathbb{Z}\times\mathbb{Z})*\mathbb{Z}$, with the complicated branching structure of the cover corresponding to the ways words can grow in that free product.