Suppose $X$ and $Y$ are schemes of finite type over a field, and let $f: X\rightarrow Y$ be a morphism. Let $\Gamma_f$ be the closed subscheme of $X\times Y$, then the first example of the FourierMukai transform says that $$f_*()=p_Y{_*}(p_X^*()\bigotimes_{\mathcal{O}_{X\times Y}}\mathcal{O}_{\Gamma_f}),$$ similarly there is an expression for $f^*$. Has anyone checked this before? I am having some difficulties (at least on some commutative algebra) in verifying them. Does anyone know why those formulas are true?

You should use the fact that $p_X^*()\otimes_{\mathcal{O}_{X\times Y}}\mathcal{O}_{\Gamma_f}\cong i_*(p_X_{\Gamma_f})^*()$ where $i:\Gamma_f\to X\times Y$ is the inclusion. Then the statment becomes much easier. 


Indeed, as follows from the comments below, maps between schemes provide examples of FourierMukai transform, most famous example being a similar map with additional twisting by a bundle in $A\times \hat A$ for an Abelian variety $A$. Anyway, since the restriction $p'_X:\Gamma_f\to X$ is actually an isomorphism (the inverse is $x\mapsto (x, f(x))$) and the composition $p_Y\circ {p'_X}^{1}: X \to \Gamma_f \to Y$ is exactly $f$, the statement you have written is actually equivalent to $f_* () = f_*()$. Thus there is no hard commutative algebra stuff. 

