At the first glance, I don't think we should solve this problem by using reflection. then I just inspect the problem, it seems the init speed is like (p, q), so after k steps, we should reach (kp, kq) without blocking. Since now we are in this square, 0 represents (p, 0), 1 represents (p, p), 2 represents (0, p) which we can rephrase 0 as x-distance is 1 more length than y-distance, 1 as x-distnace same as y-distance and 2 as y-distance is 1 more than x-distance.
Let us check point (kp, kq), if it stops which means that kq = mp. So we know that kq = mp = lcm(p, q). then we know k = lcm(p, q) / q = p / gcd(p, q) and m = lcm(p, q) / p = q / gcd(p, q), so finally we get (kp, mp). the meeting point should be (k%2, m%2)