Since we know that $ \sqrt[3]{-8}=-2 $ is a rational number, $$f(\sqrt[3]{-8})=|\lfloor{-2}\rfloor|=2.$$Continuing from here, we know that $ -\pi $ is irrational, thus $$f(-\pi)=\lceil{-\pi}\rceil^2=(-3)^2=9.$$Because 50 is not a perfect square, $ \sqrt{50} $ must be irrational as well, so $$f(\sqrt{50})=\lceil{\sqrt{50}}\rceil^2=8^2=64.$$Finally, we know that $ \frac{9}{2} $ is a rational number, so $$f\left(\frac{9}{2}\right)=\left|\left\lfloor{\frac92}\right\rfloor\right|=4.$$Therefore $$f(\sqrt[3]{-8})+f(-\pi)+f(\sqrt{50})+f\left(\frac{9}{2}\right)=2+9+64+4=\boxed{79}.$$