Homotopy equivalent spaces have homotopy equivalent universal covers

You are right that the lift of the homotopy $p\tilde{g}\tilde{f}$ starting at $ \tilde{g} \tilde{f}$ may not end at the identity. However, it does end at a lift of the identity, that is, a deck transformation $\phi$ of $\widetilde{X}$. Now, this implies that $\tilde{g}\tilde{f} \simeq \phi$ so $\phi^{-1} \tilde{g}\tilde{f} \simeq id_{\widetilde{X}}$. Similarly, there is a deck transformation $\psi$ of $\widetilde{Y}$ such that $\tilde{f}\tilde{g} \psi^{-1} \simeq id_{\widetilde{Y}}$.

Now, apply exercise 11 in chapter 0 to conclude that $\tilde{f}$ is a homotopy equivalence.

$p:(\widetilde{X},\tilde{x}_0)\to (X,x_0)$ and $q:(\widetilde{Y},\tilde{y}_0)\to (Y,y_0)$ be covering maps. $f:(X,x_0)\to(Y,y_0)$ be a homotopy equivalence, with inverse $g$. Then, $fg\simeq1_Y$ and $gf\simeq1_X$

Lift $fp:\widetilde{X}\to Y$ to $F=\widetilde{fp}:\widetilde{X}\to \widetilde{Y}$. Since $\widetilde{X}$ is simply connected, lift exists. Again by lifting, $G=\widetilde{gq}:\widetilde{Y}\to\widetilde{X}$

Now, $qFG=q\widetilde{fp}\widetilde{gq}=fp\widetilde{gq}=fgq$. So $FG$ is the (unique) lift of $fgq:\widetilde{Y}\to Y$. But, $fgq\simeq1_Yq=q$ and by homotopy lifting, $FG\simeq \tilde{q}=1_\widetilde{Y}$

Similarly, $GF\simeq 1_\tilde{X}$. Thus, $F:\widetilde{X}\to\widetilde{Y}$ is a homotopy equivalence.

While Andrew Hanlon gives in his answer what seems to me to be the correct idea (the lift is determined up to a deck transformation), here is an amusing (probably circular) alternative, just for fun.

We know that there is a lift $\tilde{f} : \tilde{X} \to \tilde{Y}$. So $\tilde{f}$ is a continuous map between two spaces, and since the map $f$ induced isomorphisms on all higher homotopy groups, being a homotopy equivalence, and since a covering map induces isomorphisms on the higher homotopy groups, it follows from the simply-connectedness of universal covers that $\tilde{f}$ induces isomorphisms on all homotopy groups.

So in the case that $X$ and $Y$ are connected CW-complexes, it follows from Whiteheads theorem that $f$ is a homotopy equivalence.