Identity Involving Lie Derivative and Local Flows I'm trying to show,
$$ \frac{d}{dt} \varphi_t^* \omega = \varphi_t^* \left( \mathcal{L}_{X_t} \omega \right)$$
but I have another question as well. Every case in which the lie derivative is mentioned, that notation $\mathcal{L}_{X_t}$ has never been given interpretation and so I would like to understand this notation first.
Given $\frac{d}{dt}\bigr|_{t=0} \varphi_t = X_p = X_{\varphi_0(p)}$ and so $\frac{d}{dt} \varphi_t = X_{\varphi_t(p)}$ which we can call $X_t$. And so,
$$\mathcal{L}_{X_t} = \mathcal{L}_{\frac{d}{dt} \varphi_t}$$
Is this correct? 
Update: attempt to prove identity
 A: In order to prove the identity:
$$\frac{\mathrm{d}}{\mathrm{d}t}{\varphi_t}^*\omega={\varphi_t}^*(\mathcal{L}_{X_t}\omega),$$


*

*Prove that it holds for functions ($0$-forms), it is just the chain rule,

*Prove that if it holds for $\omega$, then it holds for $\mathrm{d}\omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,

*Prove that if it holds for $\omega$ and $\omega'$, then it holds for $\omega\wedge\omega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,

*Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.
Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.
A: Edit: The following answer assumes $\varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.
Edit 2: It seems like OP was asking about this case, in which case my answer would apply.

If $\omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as,
$$ \mathcal{L}_X\omega|_p = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \varphi_t^* \omega|_p = \lim_{t \rightarrow 0} \frac1t\left(\varphi_t^*(\omega_{\varphi(p)}) - \omega_p \right), $$
where $\varphi_t$ is the associated flow of $X.$
From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,
$$\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=t_0} \varphi_t^*\omega = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \varphi_{t_0}^* \varphi_t^*\omega = \left.\varphi_{t_0}^* \frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \varphi_t^*\omega = \varphi_{t_0}^* \mathcal{L}_{X}\omega. $$
I'll leave you to check the details on a pointwise level.

Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define,
$$ \varphi^*\omega|_p = \varphi^*|_{\varphi(p)} \omega_{\varphi(p)} \in E_p,$$
where $E$ is whatever bundle $\omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.
In general there's nothing special about $X_t = \varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.
