Cartan's magic formula

Putting in the limit made it more mysterious. You're just using the fact that for smooth functions mixed partials are equal. That is, $$\frac{\partial}{\partial t}\frac{\partial}{\partial x^j}=\frac{\partial}{\partial x^j}\frac{\partial}{\partial t}.$$

EDIT: In particular, if you write it out in local coordinates, it's computed by taking partial derivatives with the respect to the variables $x^1,\dots,x^n$. Writing $\phi_t^*\omega=\sum f_I(x,t)dx^I$ (where $I=(i_1,\dots,i_k)$ ranges over length-$k$ multiindices), then $d\phi_t^*\omega=\sum \frac{\partial f_I(x,t)}{\partial x^j}\,dx^j\wedge dx^I$, etc.

Here is an attempt to answer my question following Ted Shifrin's comment:

First, we may write $\phi_t^* \omega(x)= \sum\limits_{I} f_I(x,t)dx^I$ where $I$ ranges over length-$k$ multiindices. Then

$$\begin{array}{ll} dL_X \omega & = d \frac{d}{d t}_{|t=0} \phi_t^* \omega = d \sum\limits_I \frac{\partial }{\partial t}_{|t=0} f_I dx^I \\ & = \sum\limits_I \sum\limits_{i=1}^n \frac{\partial}{\partial x_i} \frac{\partial}{\partial t}_{|t=0} f_I dx^i \wedge dx^I = \sum\limits_I \sum\limits_{i=1}^n \frac{\partial}{\partial t}_{|t=0} \frac{\partial}{\partial x_i} f_I dx^i \wedge dx^I \\ & = \sum\limits_I \frac{d}{d t}_{|t=0} df_I \wedge dx^I = \frac{d}{dt}_{|t=0} d \phi_t^* \omega = \frac{d}{dt}_{|t=0} \phi_t^* d \omega = L_Xd\omega \end{array}$$