First of all,
$$
\max_{x\in K}\left(\max_{y\in L}f(x,y)\right)\le\max_{(x,y)\in K\times L}f(x,y)
$$
because for each $x\in K$, we are picking one value of $y_x\in L$ on the left whereas we are taking all possible pairs on the right. That is, the set of $(x,y)$ we are maximizing over on the left is smaller than the set of $(x,y)$ on the right.

However,
$$
\max_{x\in K}\left(\max_{y\in L}f(x,y)\right)\ge\max_{(x,y)\in K\times L}f(x,y)
$$
because for the $(x_0,y_0)$ where the max is attained on the right, we have that