Getting solutions to Ricci-DeTurck flow from solutions to Ricci flow

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Let $\overline{g}(t)$ be a solution to the Ricci flow, $\tilde{g}$ be an arbitrary (fixed) background metric and denote by $\varphi_{t}$ the solution of the harmonic map heat flow

$$\frac{\partial}{\partial t} \varphi_{t}=\Delta_{\bar{g}(t), \bar{g}} \varphi$$

with respect to $\overline{g}(t)$ and $\tilde{g}$. I want to check that

$$g(t) = \left(\varphi_{t}^{-1}\right)^{*} \bar{g}(t)$$

solves Ricci-DeTurck flow. My calculations yield:

$$\begin{aligned} \frac{\partial}{\partial t} g(t) &=\frac{\partial}{\partial t}\left(\left(\varphi_{t}^{-1}\right)^{*} \bar{g}(t)\right) \\ &=\left(\varphi_{t}^{-1}\right)^{*}\left(\frac{\partial}{\partial t} \bar{g}(t)\right)+ \left(\varphi_{t}^{-1}\right)^{*}\left(\mathcal{L}_{\left(\frac{\partial}{\partial t} \varphi_{t}^{-1} \right)}\bar{g}(t)\right) \end{aligned}$$

But in the book I'm reading, their computations yield: $$\begin{aligned} \frac{\partial}{\partial t} g(t) &=\frac{\partial}{\partial t}\left(\left(\varphi_{t}^{-1}\right)^{*} \bar{g}(t)\right) \\ &=\left(\varphi_{t}^{-1}\right)^{*}\left(\frac{\partial}{\partial t} \bar{g}(t)\right)+\mathcal{L}_{\left(\varphi_{t}^{-1}\right)^{*}\left(\frac{\partial}{\partial t} \varphi_{t}\right)}\left[\left(\varphi_{t}^{-1}\right)^{*} \bar{g}(t)\right] \\ &=\left(\varphi_{t}^{-1}\right)^{*}(-2 \operatorname{Rc}[\bar{g}(t)])+\mathcal{L}_{\left(\varphi_{t}^{-1}\right)^{*}\left(\varphi_{t}^{*}[W(t)]\right)}[g(t)] \end{aligned}$$

But I'm having some trouble getting from:

$$\left(\varphi_{t}^{-1}\right)^{*}\left(\mathcal{L}_{\left(\frac{\partial}{\partial t} \varphi_{t}^{-1} \right)}\bar{g}(t)\right)$$

to $$\mathcal{L}_{\left(\varphi_{t}^{-1}\right)^{*}\left(\frac{\partial}{\partial t} \varphi_{t}\right)}\left[\left(\varphi_{t}^{-1}\right)^{*} \bar{g}(t)\right]$$

And also, my understanding is that

$$\frac{\partial}{\partial t} \varphi_{t}=\Delta_{\bar{g}(t), \bar{g}} \varphi = - W \circ \varphi$$

How could the above equal $\varphi_{t}^{*}[W(t)]$?