Let $S$ be a hyperbolic surface of finite type, i.e. $S$ is a surface of genus $g$, with $b$ boundary components and $n$ punctures with $2-2g-b-n<0$.
Fix a hyperbolic metric on $S$. Let $x$ be a non-simple closed geodesic in $S$, i.e. $x$ have self-intersections.
If $A$ and $B$ are any two geodesic lifts of $x$ to the upper half plane $\mathbb{H}$, then does $A$ and $B$ always intersect?
The answer is no: there will always be two geodesic lifts of $x$ which do not cross.
If you understand the deck transformation action of $\pi_1 S$ on $\mathbb{H}$, the proof is relatively straightforward. Here's some details of how this is done.
Let $\{X_i\}_{i \in I}$ be the set of all lifts of $x$ to $\mathbb{H}$, each of which is a geodesic line in $\mathbb{H}$; I'll call these simply lines. This set of lines in $\mathbb{H}$ is invariant under the deck transformation action of the group $\pi_1 S$ on $\mathbb{H}$.
Here are some examples of how such a set of lines might appear:
You can see in these examples that although there are many lines that do cross, there are also many lines that do not cross. In fact there must always be a pair of lines that does not cross.
To prove it, choose a line $X_i$ in the set. Let $E_i$ be the unordered pair of endpoints of $X_i$.
We may always choose a nontrivial $g \in \pi_1 S$ with corresponding deck transformation $G : \mathbb{H} \to \mathbb{H}$ so that $G$ is a "translation" having two fixed points $p^-,p^+$ on the circle at infinity, and so that $$E_i \cap \{p^-,p^+\} = \emptyset $$ (I'll show below why $g$ and $G$ can always be chosen in this manner).
A key property of $G$ is that $p^+$ is an attractor and $p^-$ is a repeller, meaning that for each point $q$ in the circle at infinity, if $q \ne p^-,p^+$ then $$\lim_{n\to\infty}G^n(q)=p^+ \quad\text{and}\quad \lim_{n\to-\infty} G^n(q)=p^- $$ Apply this to the two points $q \in E_i$, and it follows that for large $n>0$ the endpoints $E_j = G^n(E_i)$ of the line $X_j = G^n(X_i)$ are close to $p^+$ and the endpoints $E_k = G^{-n}(E_i)$ of the line $X_k = G^{-n}(X_i)$ are close to $p^-$. If they are close enough, i.e. if $n$ is big enough, it follows that $X_j$ and $X_k$ do not cross.
To choose $G$, start with three closed geodesics $c_1,c_2,c_3$ on $S$ which are pairwise transverse. Let $g_1, g_2, g_3 \in \pi_1(S)$ be elements representing the conjugacy classes associated to $c_1,c_2,c_3$. Let $G_1,G_2,G_3$ be the corresponding deck transformations. It follows that the attractor repeller pairs $\{p^+_1,p^-_1\}$, $\{p^+_2,p^-_2\}$, $\{p^+_3,p^-_3\}$ of $G_1,G_2,G_3$ are pairwise disjoint. The two-point set $E_i$ can intersect at most two of these attractor repeller pairs, and so is disjoint from at least one of those pairs.
