Sorry, I will have to begin my question by directly pasting a few snapshots. The following material comes from Geometric Relativity by Dan A. Lee.
I'd like to derive equation (2.4). What I have tried so far is that if $h_t=\Phi_t^*h$, we would obtain $h_t=\Phi_t^*g$. Since $d\Phi_t$ is an isomorphism, we should agree that $(e_i(t))$ is a local frame on $\Sigma_t$. It follows that $$\begin{align} \frac{\partial}{\partial t}(h_t)_{ij}&=\frac{\partial}{\partial t}\langle e_i(t),e_j(t)\rangle_{h_t}\\ &=\frac{\partial}{\partial t}(\Phi_t^*g)(e_i,e_j)\\ &=\frac{\partial}{\partial t}g_{\Phi_t}(e_i(t),e_j(t)). \end{align}$$ Now I wonder why equation (C) is possible. Are there any tricks? Thank you.
Edit. I got some ideas though still lost in equation (C). To arrive at equation (2.4), you can substitute $t=0$ into equation (E). To get (E), you can use equality of mixed partials to conclude the Lie bracket of $X_t$ and $e_i(t)$ is the zero vector field. Finally, equation (D) can be derived using the fact that Levi-Civita connections are metric connections.
First of all, I disagree with your first line $(h_t)_{i,j} = \left<e_i(t),e_j(t)\right>_{h_t}$ which should be $\left<e_i,e_j\right>_{h_t}$, which is what Lee wrote at equation $(A)$. Indeed, $h_t$ is a metric on $\Sigma$ and not on $\Sigma_t$. If $\iota_t : \Sigma_t \hookrightarrow M$ is your inclusion, $\iota_t^*g$ is the metric on $\Sigma_t$ induced by $g$ and $h_t = \Phi_t^*\iota_t^*g$ is a metric on $\Sigma$ (it is the defintion of $h_t$). Remember that each $(h_t)_{i,j}$ is a smooth function from an open subset of $\Sigma$ (and not $\Sigma_t$ !) in $\mathbb{R}$.
You have in fact, for all $x \in \Sigma$ where the $(e_i)$ are defined, \begin{align*} (h_t)_{i,j}(x) & = \left<e_i(x),e_j(x)\right>_{h_t(x)}\\ & = \left<e_i(x),e_j(x)\right>_{\Phi_t^*g(x)} \textrm{ where I skipped the $\iota_t^*$ by abuse of notation.}\\ & = \left<d\Phi_t(x)e_i(x),d\Phi_t(x)e_j(x)\right>_{g(\Phi_t(x))}\\ & = \left<e_i(t)(\Phi_t(x)),e_j(t)(\Phi_t(x))\right>_{g(\Phi_t(x))}\\ & = \Phi_t^*(\left<e_i(t),e_j(t)\right>_{g})(x),\\ \end{align*} which is equation $(B)$ without the $\frac{\partial}{\partial t}$ and evaluated at a point $x$.
Now, the trick is to notice that $e_i(t)$ can be extended to an open subset of $M$ and that it doesn't really depend on the time $t$ in the sense that $\frac{\partial e_i}{\partial t}(t) = 0$ when we are on $\Sigma_t$. Indeed, take one point $x_0 \in \Sigma$ and extend each $e_i$ in an open neighborhood of $x_0$ (it is always possible to do that) and let extend $e_i(t)$ with $e_i(t)(y) = d\Phi_t(\Phi_t^{-1}(y))e_i(\Phi_t^{-1}(y))$ which exists in a neighborhood of $\Phi_t(x_0) \in \Sigma_t$ and extends $e_i$.
Therefore, $t,y \mapsto e_i(t)(y)$ exists in an open subset of $]-\epsilon,\epsilon[ \times M$ so $\frac{\partial e_i}{\partial t}$ makes sens. Now, we have by definition that for $x$ close enough to $x_0$, $$ e_i(t)(\Phi_t(x)) - d\Phi_t(x)e_i(x) = 0 $$ thus when we differentiate with respect to $t$, \begin{align*} 0 & = \frac{\partial e_i}{\partial t}(t)(\Phi_t(x)) + de_i(t)(\Phi_t(x))\frac{\partial}{\partial t}\Phi_t(x) - d\frac{\partial}{\partial t}\Phi_t(x)e_i(x)\\ & = \frac{\partial e_i}{\partial t}(t)(\Phi_t(x)) + de_i(t)(\Phi_t(x))X_t(\Phi_t(x)) -d(X_t \circ \Phi_t)(x)e_i(x)\\ & = \frac{\partial e_i}{\partial t}(t)(\Phi_t(x)) + de_i(t)(\Phi_t(x))X_t(\Phi_t(x)) - dX_t(\Phi_t(x))d\Phi_t(x)e_i(x)\\ & = \frac{\partial e_i}{\partial t}(t)(\Phi_t(x)) + de_i(t)(\Phi_t(x))X_t(\Phi_t(x)) - dX_t(\Phi_t(x))e_i(t)(\Phi_t(x))\\ & = \left(\frac{\partial e_i}{\partial t}(t) - [X_t,e_i(t)]\right)(\Phi_t(x)). \end{align*} We deduce that $\frac{\partial e_i}{\partial t}(t) = [X_t,e_i(t)]$ when it is well defined. In particular, on $\Sigma_t$, $[X_t,e_i(t)] = 0$ (I didn't verify it but you said you proved it) so $\frac{\partial e_i}{\partial t} = 0$ (notice that we can't conclude that $e_i$ is locally constant with repect to time since it only works on $\Sigma_t$, not on the open set where the $e_i(t)$ have been continued).
Now, we have the tools to conclude. When we are close enough to $x_0$, \begin{align*} \frac{\partial}{\partial t}\Phi_t^*\left<e_i(t),e_j(t)\right>_g & = \Phi_t^*\frac{\partial}{\partial t}\left<e_i(t),e_j(t)\right>_g + \Phi_t^*(d\left<e_i(t),e_j(t)\right>_g)\frac{\partial}{\partial t}\Phi_t(x)\\ & = \Phi_t^*\left(\left<\frac{\partial e_i}{\partial t}(t),e_j(t)\right>_g + \left<e_i(t),\frac{\partial e_j}{\partial t}(t)\right>_g\right) + \Phi_t^*(d\left<e_i(t),e_j(t)\right>_g)\Phi_t^*X_t\\ & = \Phi_t^*\left(\left<\frac{\partial e_i}{\partial t}(t),e_j(t)\right>_g + \left<e_i(t),\frac{\partial e_j}{\partial t}(t)\right>_g + d\left<e_i(t),e_j(t)\right>_gX_t\right)\\ & = \Phi_t^*\left(\left<\frac{\partial e_i}{\partial t}(t),e_j(t)\right>_g + \left<e_i(t),\frac{\partial e_j}{\partial t}(t)\right>_g + X_t\left<e_i(t),e_j(t)\right>_g\right). \end{align*} And at a point $x \in \Sigma$, we have $\Phi_t(x) \in \Sigma_t$ so the $\frac{\partial e_i}{\partial t}$ vanish and, $$ \frac{\partial}{\partial t}(h_t)_{i,j} = \Phi_t^*(X_t\left<e_i(t),e_j(t)\right>_g). $$ It is equation $(C)$. I don't know however if it is the fastest method, maybe you can prove $(C)$ without extending the $e_i$ but you will struggle to compute the derivative with respect to time since the domain of definition of $e_i(t)$ changes when $t$ changes.