Subdivide the universal covering space $\mathbb R^2$ in the usual manner as squares, with vertical lines $x=m$ and horizontal lines $y=n$ for integers $m,n \in \mathbb Z$.
To visualize the desired covering space, draw two vectors based at the origin: $v = \langle 3,0 \rangle$ corresponding to $a^3$; and $w = \langle 2,1 \rangle$ corresponding to $a^2 b$. Let $P$ be the parallelogram determined by the vectors $v,w$, which form two sides of $P$, the other two sides then being determined. Now glue opposite sides of $P$ to form a quotient space $S$. And as usual, gluing opposite sides of the unit square $Q = [0,1] \times [0,1]$ gives the covering space $T$.
You can then use the pattern of intersections of the parallelograph $P$ with unit squares $[m,m+1] \times [n,n+1]$ to define the desired covering map $Q \mapsto T$.
It is theoretically more straightforward to view this construction using orbit spaces of deck transformations. If I have $(a,b) \in \mathbb R^2$ let me use $\tau_{(a,b)}$ to represent the translation $\tau_{a,b}(x,y) (x+a,y+b)$. Thus we can think of $T$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(1,0)},\tau_{(0,1)} \rangle$ (with fundamental domain $[0,1] \times [0,1]$), so
$$T = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle
$$
and we can think of $S$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(3,0)}, \tau_{(2,1)} \rangle$ (with fundamental domain $P$), so
$$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle
$$
This way, the desired quotient map $S \mapsto T$ can be precisely defined as the map
$$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle \mapsto = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle = T
$$
that is induced by the identity map $\mathbb R^2 \mapsto \mathbb R^2$.
Since isomorphism classes of covering spaces correspond to conjugacy classes of subgroups, it suffices to show that the two subgroups are not conjugate. This is easy in this case, because $\mathbb{Z}^2$ is abelian.
Edit: To be more explicit, $\mathbb{Z}^2$ is abelian, so for any $g \in \mathbb{Z}^2$, $gBg^{-1} = Bgg^{-1} = B$, so you just have to show that $B \neq K$. Clearly, $(3,1) \in B$. If we assume $(3,1) \in K$, we know
$$a(3,0) + b(0,2) = (3,1)$$
for integers $a,b$. But this is impossible. Then, $(3,1) \not \in K$, so $K \neq B$, meaning $gBg^{-1} \neq K$ for all $g$.
Best Answer
We want to show that the only covering spaces of the $2$-dimensional torus are tori, cylinders and the plane. We know that $\pi_1(S^1\times S^1)=\mathbb{Z}\times \mathbb{Z}$, which has the following subgroups: 1. the trivial subgroup, 2. free abelian groups with one generator $(p, q)$, 3. free abelian groups with two generators $(p, q)$ and $(r, s)$ such that $ps − qr\neq 0$. For each subgroup one constructs now a covering space. If the subgroup is trivial we obtain the universal covering space $\mathbb{R}^2$ with covering map $p : \mathbb{R}^2\rightarrow S_1 × S_1$, $$p(x, y) = (e^{2πix}, e^{2πiy}).$$ For the second case we obtain $S^1\times \mathbb{R}$, and for the third case $S^1\times S^1$. As you said, only the third one is compact.