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

Letting now 
$$\th_j:=\arctan m_j$$
and using the rotational symmetry of the distribution of $(U,V)$, we get 
$$p_X(t_0,\ep)=\frac{\th_2-\th_1}{\pi}.
$$