$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, 
$$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$
where 
$$m_1:=\min_{t\in[t_0-\ep,t_0+\ep]}r(t),\quad 
m_2:=\max_{t\in[t_0-\ep,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$
this follows because any continuous function maps any compact interval onto a compact interval. 

Letting now 
$$\th_j:=\arctan m_j$$
and using the rotational symmetry of the distribution of $(U,V)$, we get 
$$p_X(t_0,\ep)=
\left\{
\begin{aligned}
\frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ 
\frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0.
\end{aligned}
\right.
$$