Let $\Gamma \le \operatorname{PSL}(2,\mathbb{R})$ be a discrete subgroup and $\Sigma:=\mathbb{D}^2/\Gamma$ be the quotient. Then $\Sigma$ is a hyperbolic surface whose universal cover is $\mathbb{D}^2$ and the fundamental group $\pi_1(\Sigma,x)$ ($x\in\Sigma$) is isomorphic to $\Gamma$. Fix a point $\tilde{x}\in\mathbb{D}^2$ over $x$.
For two loops $\alpha, \beta:S^1(\cong \mathbb{R}/\mathbb{Z})\rightarrow \Sigma$ in $\pi_1(\Sigma,x)$ (so $\alpha(0)=\alpha(1)=\beta(0)=\beta(1)=x$), let
$\tilde{\alpha}_0, \tilde{\beta}_0:\mathbb{R}\rightarrow\mathbb{D}^2$ be the liftings of $\alpha, \beta$, respectively, with $\tilde{\alpha}_0(0)=\tilde{\beta}_0(0)=\tilde{x}$.
Then let
$\tilde{\alpha}_1:\mathbb{R}\rightarrow \mathbb{D}^2$ be the lifting of $\alpha$ with $\tilde{\alpha}_1(0)=\tilde{\beta}_0(1)$,
and
$\tilde{\beta}_1:\mathbb{R}\rightarrow \mathbb{D}^2$ the lifting of $\beta$ with $\tilde{\beta}_1(0)=\tilde{\alpha}_0(1)$.
Now I want to show that $\tilde{\alpha}_0$, $\tilde{\beta}_0$, $ \tilde{\alpha}_1$ and $\tilde{\beta}_1$ don't make an embedded quadrangle in $\mathbb{D}^2$. Can I show this using techniques in hyperbolic geometry or find some counterexamples? Any help or hint will be appreciated.
Meanwhile I found a counterexample on my own on a punctured torus, using @LeeMosher 's comment.