$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0,t_0+\ep]}r(t),\quad m_2:=\max_{t\in[t_0,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ 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 $$p_X(t_0,\ep)= \left\{ \begin{aligned} \frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ \frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0. \end{aligned} \right. $$
