Let $M$ be a smooth manifold. Suppose $V:J\times M\to M$ is a smooth time-dependent vector field and $\psi:\mathcal{E}\to M $ is its time-dependent flow. For any smooth function $f\in C^{\infty}(M)$ and any $(t,t_0,p)\in \mathcal{E}$ how to show that $$\dfrac{d}{dt}\bigg|_{t=t_0} \left(\psi_{t,t_0}^* df\right)_p = d\left(\mathcal{L}_{V_{t_0}}f\right)_p?$$ where $\psi_{t,t_0}$ is just $\psi$ with $t$ and $t_0$ fixed, $V_{t_0}$ is $V$ with $t_0$ fixed and $\mathcal{L}_{V_{t_0}}f$ is the Lie derivative of $f$ with respect to $V_{t_0}$.
There is a proof of this in Lee's book Introduction to Smooth Manifolds, 2nd edition on page 572 by showing that the operators $d$ and $\frac{\partial}{\partial t}$ commutes. His arguments is-"In any smooth local coordinates $(x_i)$, the function $\psi_{t,t_0}^* f(x) = f(\psi(t,t_0,x))$ depends smoothly on all $n+1$ variables $(t,x_1,\cdots, x_n)$. Thus, the operator $\frac{d}{dt}$ (which is more properly written as $\frac{\partial}{\partial t}$ in this situation) commutes with each of the partial derivatives $\frac{\partial}{\partial x^i}$ when applied to $\psi_{t,t_0}^* f$. In particular, this means that the exterior derivative operator $d$ commutes with $\frac{\partial}{\partial t}$ , and so $$\dfrac{d}{dt}\bigg|_{t=t_0} \left(\psi_{t,t_0}^* df\right)_p = \dfrac{\partial}{\partial t}\bigg|_{t_0} d \left(\psi_{t,t_0}^*f\right)_p = d \left(\dfrac{\partial}{\partial t}\bigg|_{t_0}\left(\psi_{t,t_0}^*f\right)\right)_p = d\left(\mathcal{L}_{V_{t_0}}f\right)_p.$$
What I dont really understand is the part where he interchange the operator $d$ and $\frac{\partial}{\partial t}$, can anyone write explicitly how to interchange these two operators.