In page 135 of Ethier and Kurtz(1986 - Markov processes, convergence and characterization) one reads:
The question is, do we need the factor $a_\beta$ to be squared?
Since we have (8.22) it seems that can we use triangle inequality as follows: $$ \sup_{0\leq v\leq 2\delta \wedge \tau} q^\beta(X(\tau), X(\tau - v)) \\ \leq \sup_{0\leq v\leq 2\delta \wedge \tau} \bigg(q^\beta(X(\tau +U ), X(\tau - v)) + q^\beta(X(\tau +U ), X(\tau))\bigg)$$
To get rid of the square on the factor $a_\beta$?
No you can't,
note that $q(a,b) \leq q(a,c) + q(c,b)$ but when you take powers,
$$q^\beta(a,b) = \big(q(a,c) + q(c,b)\big)^\beta \leq 2^{\beta-1} (q^\beta(a,c) + q^\beta (c,b)) $$