They claim later on in the proof that $G’(0)=0$. I don’t see this. I would say that we have $$ G(t)=\int_0^tF(\tau)d\tau-\int_0^{-t}F(\tau)d\tau-t/3[F(-t)+4F(0)+F(t)], $$ which yields $$ G’(0)=F(0)-F(0)-1/3[6\cdot F(0)]=-2F(0). $$ Could someone explain what is going on?
You are missing the chain rule for the derivative of the second term. If we define $I(t):=\int_0^{t}F(\tau)\,d\tau$, then $I'(t)=F(t)$, and therefore $$ \frac{d}{dt}\left[\int_0^{-t}F(\tau)\,d\tau\right]=\frac{d}{dt}\left[I(-t)\right]=-I'(-t)=-F(-t). $$ So, we get $$ G'(0)=F(0)-(-F(-0))-\frac{1}{3}\left[6F(0)\right]=F(0)+F(0)-2F(0)=0. $$