This is just a partial answer addressing the continuity: $U$ is continuous, even uniformly continuous. Below is the idea.

Denote $\tau_x^y = \inf\{ t\ge 0: X_t = y\,|\, X_0 = x\}$, then $\tau_x^y<\infty$ a.s. in view of the uniform non-degeneracy of $v$ and boundedness of $m$, so, using the strong Markov property, $$U(x) = rE\left[\int_0^{\tau_x^y} e^{-rs}f(X_t)dt \,\Big|\, X_0=x\right] + E\big[e^{-r\tau_x^y}\big]U(y).$$
The first term converges to $0$ uniformly as $|y-x|\to 0$ in view of the uniform non-degeneracy of $v$ and boundedness of $m$. For the same reason, $E\big[e^{-r\tau_x^y}\big]\to 1$ uniformly.

Update: $U'$ is in general discontinuous. Start by noting that we expect $U$ to solve equation $$rU - \mathcal A U - f = 0,\tag{1}$$ where $\mathcal A$ is the generator of $X$, at least in some weak sense. Then it is already clear that $U'$ or $U''$ must be discontinuous in the point where $f$, $m$ or $v$ are discontinuous.

But $U'$ may be discontinuous even if $f$ and $m$ are continuous. Consider a simple situation: $m = 0$, $f$ smooth, $v(x) = \sigma_1 \mathbf{1}_{x\ge 0} + \sigma_2 \mathbf{1}_{x<0}$. In this case $X_t = W_t\big( \sigma_1\mathbf{1}_{W_t\ge 0} + \sigma_2 \mathbf{1}_{W_t<0}\big)$, where $W$ is a Wiener process. Then
$$
T_t g(x) = E_x[g\big(W_t( \sigma_1\mathbf{1}_{W_t\ge 0} + \sigma_2 \mathbf{1}_{W_t<0})\big)] = P_t h(x),
$$
where $P$ is the semigroup of $W$ (the heat semigroup), and $h(x) = g(x v(x))$. Then $\mathcal A g(x) = \frac12 h''(x)$, so it is easy to see that the domain of $\mathcal A$ in $C_b(\mathbb R)$ consists of twice continuously differentiable functions $g$ on $\mathbb R\setminus\{0\}$ such that $g'(0+)\sigma_1 = g'(0-)\sigma_2$ and $g''(0+)\sigma_1^2 = g''(0-)\sigma_2^2$.

In other words, $U'$ should have jumps where $v$ does. It is even more obvious that it jumps where $m$ does; when both jump, it will be more complicated.