Why commutator of positive and negative parts of scalar field is equal to the Feynman propagator?

Just evaluate the $p^0$ part of the integral. For $x^0>y^0$, we can close the contour below and enclose the pole at $p^0=E_mathbf{p}$ (there's a pole at this point because $p$ is on-shell, so $p^2-m^2=0$). Then we evaluate the integral using Cauchy's integral formula, (I'm playing fast and loose with the $iepsilon$) $$int frac{d p^0}{2 pi } frac{ie^{-i p cdot(x-y)}}{p^{2}-m^{2}+i epsilon}=intfrac{d p^0}{2 pi i} frac{-e^{-ip cdot(x-y)}}{(p^0+sqrt{m^{2}+mathbf{p}^2})(p^0-sqrt{m^{2}+mathbf{p}^2})+i epsilon}$$ $$=-intfrac{d p^0}{2 pi i} frac{e^{-ip cdot(x-y)}}{(p^0+E_mathbf{p})(p^0-E_mathbf{p})+i epsilon}=frac{e^{-ip cdot(x-y)}}{2E_mathbf{p}+i epsilon},$$ where it's implicitly understood that $p^0=E_mathbf{p}$ in the final $e^{-ip cdot(x-y)}$. Note that a sign flip occurred because we're integrating clockwise instead of counterclockwise as assumed in Cauchy's integral formula. Using the above, we get (for $x^0>y^0$ only) that $$int frac{d^{4} p}{(2 pi)^{4}} frac{i e^{-i p cdot(x-y)}}{p^{2}-m^{2}+i epsilon}=int frac{d^{3} p}{(2 pi)^{3}} frac{e^{-i p cdot(x-y)}}{2E_mathbf{p}+i epsilon}.$$