If $M$ is a manifold and $\hat M$ its universal cover, $\hat M$ is endowed with a differentiable structure such that $p:\hat M\rightarrow M$ is differentiable. For every $x\in M$, there exists a neighborhood $U$ of $x$ such that $U$ is a domain chart $\phi:U\rightarrow\mathbb{R}^n$ and $p^{-1}(U)=\bigcup V_i$ and $p_{\mid V_i}\rightarrow U$ is an homeomorphism, the differentiable structure is defined by supposing that $p_i$ is a diffeomorphism which induces the chart $\phi\circ p_i$ on $\hat V_i$.
Now look at the action of an element $\gamma\in\pi_1(M)$ on $\hat M$, i.e. a covering transformation. We will prove that this action is already smooth. Let $p(y)=x$. There exist $i$ and $j$ such that $y\in V_i$ and $\gamma . y\in V_j$,
However, the map $(\phi\circ p_j) \circ \gamma \circ (\phi\circ p_i)^{-1}$ is the identity, this implies that action of $\gamma$ is differentiable.
This proves that every covering transformation is smooth.
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 $\mathbb R^2 \mapsto \mathbb R^2$.
Best Answer
let me use a basic example : the covering $\mathbb R \to \mathbb S^1 = \mathbb R/\mathbb Z$ (simply given by the canonical projection). In some sense, even if $\mathbb R$ is "bigger" than $\mathbb S^1$, it is "simpler" for the fondamental group : $1 \to \mathbb Z$ is, indeed, injective.
More generally, if I denote $\widetilde X$ the universal covering of $X$, then, using the usual order of coverings, $X \leq E \leq \widetilde X$, so $E$ is closer to $\widetilde X$ than $X$ is.