Let $x^0 > y^0$ where the zero above means the time component of four-vectors. This only means that the points $x,y$ in space-time are not occuring simultaneously. Then the equation is
$$\int \frac{d^3p }{(2 \pi )^3}\frac{1}{2E_p}\left(e^{-ip(x-y)}-e^{+ip(x-y)}\right) = \int \frac{d^3p }{(2 \pi )^3} \left \{ \frac{1}{2 E_p}e^{-ip(x-y)}\Bigg\vert_{p^0 = E_p} + \frac{1}{-2E_p}e^{-ip(x-y)}\Bigg \vert_{p^0=-E_p}\right \} $$
How he go from the left-side of the equation to the right-side of the equation?
I think that the first part of the right-side is clear, so we have that nothing changed. The other part was such that we performed a change of variables $\vec{p} \to -\vec p $. The integration is over all momentum space so the integral does not change and inside we get that
$$p(x-y) \to p^0(x^0-y^0) + \vec p (\vec x - \vec y) = -p(x-y) $$
where we get the equallity only if $p^0 = -E_p$. Then he continues the equallity with
$$= \int \frac{d^3p }{(2 \pi )^3}\int \frac{dp^0}{2 \pi i}\frac{-1}{p^2-m^2}e^{-ip(x-y)}$$
where I think that he used the residue theorem backwards. But this is also not so clear. We have that $E_p = \sqrt{p^2 + m^2}$ so we have poles at $p = \pm im$ how did he get the $i/(p^2-m^2)$ term of the last equality?
The $\pm im$ branch points don't matter here... they would be relevant to doing the $d^3p$ integral, but not here when we're going backwards. The relevant poles are the ones at $$ p^0 =\pm E_{\vec p}-i\epsilon$$ for the $p^0$ integral. We have added the $-i\epsilon$ to move the poles slightly below the line of integration. (This is the sense in which the integral on the RHS must be understood. This "pole prescription" yields the retarded version of the propagator, which is the one you are calculating.) With the poles on this side of the real $p_0$ line, we see that when we close the contour below (as we must do because of the $e^{-ip_0(x_0-y_0)}$ factor), we can apply the residue theorem to yield the left-hand side.