Intuition for the commutation of the pullback and the exterior derivative

Question

I know that there are already several questions to this topic but I haven‘t seen a satisfactory answer. Deriving that the pullback and the exterior derivative commute is no problem but I want to know what is the (geometric) intuition behind that.

Answer 1

What about Stokes' Theorem? $$ \int_M dF^* \omega = \int_{\partial M} F^*\omega = \int_{F(\partial M)} \omega $$ On the other hand $$ \int_M F^* d\omega = \int_{F(M)} d\omega = \int_{\partial F(M)} \omega $$ Is it easier to accept that integrating over $F(\partial M)$ is the same as integrating over $\partial F(M)$? I would argue it's easier to imagine what is going on geometrically, especially if we work with the chain complex version of Stokes' Theorem.