In order to show that the function $f(t,y_1,y_2) = (y_2, At - B\sin y_1)$, is Lipschitz-continuous with respect to the variable $\underline{y}=(y_1,y_2)$, it is sufficient to show there exists a number $K\geq 0$, such that $\vert \partial_{j} f(t,y_1,y_2)\vert\leq K$ for all $j = (y_1,y_2)$. So this one $K$ must bound each of the partials of $f$. Because $$\vert \partial_{1} f(t,y_1,y_2)\vert=\vert (0,-B\cos y_1)\vert\leq \vert B\vert$$
and $$\vert \partial_{2} f(t,y_1,y_2)\vert=\vert (1,0)\vert\leq1$$
then choosing $K:=\vert B\vert+1$ implies that the function is lipschitz-continuous with respect to $\underline{y}=(y_1,y_2)$. This $K$ is not necessarily the minimal choice, but it has no bearing on the conclusion, correct?
That is correct. Finding a bounding constant is enough to prove Lipschitz continuity. No need to find the « best » constant.
I imagine that the background of the question is differential equations.