Lemma: Given the commutative diagram
$$\begin{array}{ccccccccc} \widetilde{X} & \\
\downarrow{\small{p}} & {\searrow}^{q} \\
X_1 & \!\!\!\!\! \xleftarrow{p_1} & \!\!\!\! X_2\end{array}$$
where $p_1, p$ are covering maps, then so is $q$, where $X_1, X_2, \tilde{X}$ are all path-connected and locally path-connected.
Proof:
$q$ is surjective: $\sigma$ be a path in $X_2$ from $x_0$ and $x$. Pushforward by $p_1$ to get a path $p_1 \circ \sigma$ in $X_1$ from $p_1(x_0) = x_0'$ to $p_1(x)$. Lift to $\widetilde{X}$ to get a path $\widetilde{\sigma}$ starting from some point $x_0''$ in the fiber over $x_0'$. Pushforward by $q$ to get path $q \circ \widetilde{\sigma}$ starting at $x_0$. Uniqueness of path-lifing says $q \circ \widetilde{\sigma} \simeq \sigma$, so that $q$ maps the endpoint of $\tilde{\sigma}$ to the endpoint $x$ of $\sigma$. As $X_2$ is path connected, we can apply this argument for all $x \in X_2$ to prove $q$ is surjective.
$q$ is a covering map: Pick $x \in X_2$. Pushforward by $p_1$ to get $p_1(x)$ in $X_1$. There is a path-connected neighborhood $\mathscr{U}$ of $p_1(x)$ evenly covered by $p_1$ and $p$ (take neighborhoods evenly covered by $p_1$ and $p$ and take intersection). $\mathscr{V}$ be the slice in $p_1^{-1}(\mathscr{U})$ containing $x$. $\{\mathscr{U}_\alpha\}$ be the slices in $p^{-1}(\mathscr{U})$. $q$ maps each slice $\mathscr{U}_\alpha$ to distinct slices in ${p_{1}}^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ is then union of slices in $\{\mathscr{U}_\alpha\}$ which are mapped homeomorphically onto $\mathscr{V}$. I claim all $\mathscr{U}_\alpha$ are mapped homeomorphically on $\mathscr{V}$ by $q$. This can be proved slicewise, recalling that given a commutative diagram with any two arrows as homeomorphism, so is the third. $\blacksquare$
If $\widetilde{X}$ is simply connected, $p : \widetilde{X}\to X$ the universal cover, $p_1 : X_2 \to X_1$ a covering map, then as $p_*(\pi_1(\widetilde{X}))$ fits inside ${p_1}_*(\pi_1(X_2))$, being the trivial group, we can lift $p$ to $\tilde{p} : \widetilde{X} \to X_1$. By previous discussion, $\tilde{p}$ is a covering map, since it fits inside a commutative diagram like above. Thus, $\widetilde{X}$ covers $X_2$, as desired.
I know this is an old question but I'm currently learning this material myself and I think answering would be good practice for me. Hopefully someone will find it helpful...
First, we can employ Hatcher's 1.36 as you suggest, which is not so difficult to put into practice once we understand the universal cover. In this case, we claim the universal cover is the 4-valent tree, ie the infinite tree in which each vertex has exactly one incoming and outgoing $a$ edge, and one incoming and outgoing $b$ edge, and the incoming and outgoing edges of each type are distinct. (https://en.wikipedia.org/wiki/Cayley_graph#/media/File:Cayley_graph_of_F2.svg)
By the nature of how we've defined the edges, there is a local homeomorphism from this space to our base space. In the case of graphs, a local homeomorphism is equivalent to a covering map. Furthermore, it is a universal cover since it is simply-connected. Now to construct the covering space corresponding to the given subgroup, for $[\gamma], [\gamma'] \in \tilde{X}$, we say $[\gamma] \sim [\gamma']$ if $\gamma(1) = \gamma'(1)$ and $[\gamma \overline{\gamma}'] \in H$. The first property is automatically satisfied in this case since $X$ only has one vertex so any paths $\gamma, \gamma'$ will have $\gamma(1) = \gamma'(1)$. The second condition is just saying that when these paths are traversed as a loop (first $\gamma$ then $\gamma'$ in reverse), the homotopy class of this loop should lie in $H$. In practice, it is enough to identify the loops which correspond to the generators of $H$, hence we identify all $[\gamma]\sim[\gamma']$ such that $[\gamma \overline{\gamma}'] \in \{a^2, b^2, (ab)^4\}$. You can verify that this amounts to identifying the points $[a] \sim [a^{-1}]$, $[b] \sim [b^{-1}]$, and $[(ab)^2] \sim [(ab)^{-2}]$, where $[a]$ refers to the equivalence class represented by the closed loop in $X$ obtained by traversing the edge $a$, and $[a^{-1}]$ refers to traversing this path in reverse. Identifying these points in $\tilde{X}$ yields the covering space:
where single arrows are $a$ edges and double arrows are $b$ edges. It's clear that this is in fact a covering space and its fundamental group is generated by $\{a^2, b^2, (ab)^4\} \subset \left\langle a,b \right\rangle \cong \pi_1(X)$. Alternatively, another way to think about this is by group actions. By Hatcher's 1.39, any normal covering $\hat{X}$ corresponding to this subgroup will satisfy $$G(\hat{X}) \cong \pi(X)/\left\langle a^2, b^2, (ab)^4 \right\rangle = \left\langle a,b | a^2, b^2, (ab)^4 \right\rangle,$$ where $G(\hat{X})$ corresponds to the group of covering space automorphisms (deck transformations) of $\rho: \hat{X} \rightarrow X$. This group acts freely on the fibers $\rho^{-1}(x)$ of the cover, and since $X$ only has one vertex, the group will act freely on the vertices of $\hat{X}$. Moreover, since our subgroup is normal, the cover $\rho: \hat{X} \rightarrow X$ is regular, so the group of deck transformations acts transitively on the vertices of $\hat{X}$. To conclude, since the action is transitive and free, you can convince yourself that the vertices of $\hat{X}$ correspond to the elements of $G(\hat{X})$, and so we can identify $\hat{X}$ with the Cayley graph of $G(\hat{X})$. With $G(\hat{X}) \cong \left\langle a,b | a^2, b^2, (ab)^4 \right\rangle$, the Cayley graph is simple to draw. Notice that this method of obtaining the covering space $\hat{X}$ agrees with the first method!
Best Answer
Define the function $$\begin{align*} \varphi:&I \to \tilde X\\ &t\mapsto [\gamma_t] \end{align*}$$ Let $\gamma:I \to X$ be a path. Then, by compactness, we can find $0=t_0<t_1<\ldots < t_n =1$ such that $\gamma\left([t_i-1,t_i]\right) \subseteq U_i$ for some $U_i\in \mathcal U$, for all $i=1,\ldots,n$ where $\mathcal U$ is the basis of $X$ described at page 64.
Let $p_i$ be the homeomorphism $p:{U_i}_{[\gamma_{t_i}]}\to U_i$, the existence of which follows from the fact that $\gamma_{t_i}(1)\in U_i$ and the preceding paragraphs in Hatcher.
Claim: $\varphi=p_i^{-1}\circ \gamma$ on each $[t_{i-1},t_i]$.
Proof of claim: Let $t\in [t_{i-1},t_i]$. Then $\gamma_t(1)=\gamma(t)\in U_i$ and so ${U_i}_{[\gamma_t]}$ is well-defined. Now, $\gamma_t$ is homotopic to $\gamma_{t_i} \cdot \alpha$, where $\alpha$ is the path $s\mapsto \gamma((1-s)\cdot t_i + s\cdot t)$ which lies entirely in $U_i$. This shows that $[\gamma_t]=[\gamma_{t_i}\cdot\alpha]\in {U_i}_{[\gamma_{t_i}]}$, and thus $ {U_i}_{[\gamma_{t_i}]}={U_i}_{[\gamma_{t}]}$ by a preceding remark. Thus,
$$\varphi(t)=[\gamma_t]\in {U_i}_{[\gamma_{t_i}]}\Rightarrow (p_i\circ \varphi)(t) = \gamma_t(1)=\gamma(t)\\ \Rightarrow \varphi(t)=(p_i^{-1}\circ\gamma)(t)$$
This proves the claim. $\square$
Then the result follows by the pasting lemma.