 Solution

Let $\tilde{\alpha }$ and $\tilde{\beta }$ be lifts to the compactified hyperbolic plane $\overline{{\mathbb{H}}^2}$. Suppose they have two common points. Choose two consecutive common points along one lift. If both lie in ${\mathbb{H}}^2$, the intervening subarcs contain an innermost embedded disc. The covering projection is injective on its interior; otherwise a nontrivial deck translate would produce a still smaller such disc. Its projection is a bigon between $\alpha$ and $\beta$, contrary to the hypothesis.

The same innermost-disc argument works when one corner is on the circle at infinity, after deleting a sufficiently small horoball about that corner. The only possible obstruction would identify the ideal corner by both a hyperbolic deck transformation associated with an essential return and a parabolic deck transformation stabilizing the relevant puncture. A hyperbolic and a parabolic element of the surface group cannot have that fixed point in common, so the truncated disc again projects to a bigon. If both common points are ideal, truncate at both ends and apply the same argument. Therefore any pair of lifts intersects in at most one point of $\overline{{\mathbb{H}}^2}$.