[proofplan]
Differentiate the pulled-back metric $g(t)=\varphi_t^*\hat{g}(t)$ using the time-dependent pullback formula. The velocity field of the diffeomorphisms is defined by
\begin{align*}
V_t=\dot{\varphi}_t\circ\varphi_t^{-1}.
\end{align*}
Here $W_t \in \mathfrak{X}(M)$ denotes the DeTurck vector field associated to the evolving metric $\hat{g}(t)$ and the fixed background metric $\bar{g}$. For any smooth metric $h$ on $M$, let $\Gamma(h)^k_{ij}$ denote the Christoffel symbol of the Levi-Civita connection of $h$ in the chosen coordinate chart. Then, in any coordinate chart, $W_t$ is the vector field whose contravariant components are
\begin{align*}
(W_t)^k=\sum_{i,j}(\hat{g}(t))^{ij}\left(\Gamma(\hat{g}(t))^k_{ij}-\Gamma(\bar{g})^k_{ij}\right).
\end{align*}
The DeTurck diffeomorphism equation gives
\begin{align*}
V_t=-W_t.
\end{align*}
Thus the Lie derivative term produced by differentiating the pullback cancels the DeTurck Lie derivative term in the Ricci-DeTurck equation. Finally, Ricci curvature is natural under diffeomorphism pullback, so the pulled-back Ricci tensor is the Ricci tensor of the pulled-back metric.
[/proofplan]
[step:Identify the velocity field of the DeTurck diffeomorphisms]
For each $t \in [0,T)$, let $W_t \in \mathfrak{X}(M)$ denote the DeTurck vector field associated to $\hat{g}(t)$ and $\bar{g}$. For any smooth metric $h$ on $M$, let $\Gamma(h)^k_{ij}$ denote the Christoffel symbol of the Levi-Civita connection of $h$ in the chosen coordinate chart. Then in every coordinate chart the contravariant components of $W_t$ are
\begin{align*}
(W_t)^k=\sum_{i,j}(\hat{g}(t))^{ij}\left(\Gamma(\hat{g}(t))^k_{ij}-\Gamma(\bar{g})^k_{ij}\right).
\end{align*}
Define the time-dependent velocity field $V_t: M \to TM$ by sending each $q \in M$ to
\begin{align*}
V_t(q)=\frac{d}{dt}\varphi_t(\varphi_t^{-1}(q)).
\end{align*}
Since $\varphi_t$ satisfies
\begin{align*}
\frac{d}{dt}\varphi_t(p)=-W_t(\varphi_t(p)),
\end{align*}
putting $p=\varphi_t^{-1}(q)$ gives
\begin{align*}
V_t(q)=-W_t(q).
\end{align*}
Thus $V_t=-W_t$ as smooth vector fields on $M$. The algebraic cancellation below only uses the assumed existence of the smooth family $(\varphi_t)_{t \in [0,T)}$ of DeTurck diffeomorphisms on the interval under consideration.
[/step]
[step:Differentiate the pulled-back metric]
Let $S^2T^*M$ denote the bundle of smooth symmetric covariant $2$-tensors on $M$. For a smooth time-dependent covariant $2$-tensor $h(t)$ on $M$ and a smooth family of diffeomorphisms $\varphi_t$ with velocity field $V_t=\dot{\varphi}_t\circ\varphi_t^{-1}$, the time-dependent identity for the [pullback](/page/Pullback) of tensor fields is
\begin{align*}
\frac{d}{dt}\varphi_t^*h(t)=\varphi_t^*\left(\partial_t h(t)+\mathcal{L}_{V_t}h(t)\right).
\end{align*}
Applying this identity to the smooth map $h: M \times [0,T) \to S^2T^*M$ defined by $h(p,t)=\hat{g}(t)_p$ gives
\begin{align*}
\partial_t g(t)=\frac{d}{dt}\varphi_t^*\hat{g}(t).
\end{align*}
The pullback identity therefore yields
\begin{align*}
\partial_t g(t)=\varphi_t^*\left(\partial_t\hat{g}(t)+\mathcal{L}_{V_t}\hat{g}(t)\right).
\end{align*}
Using $V_t=-W_t$ and linearity of the Lie derivative in the vector field,
\begin{align*}
\partial_t g(t)
=\varphi_t^*\left(\partial_t\hat{g}(t)-\mathcal{L}_{W_t}\hat{g}(t)\right).
\end{align*}
[guided]
We must be careful here because both objects in $g(t)=\varphi_t^*\hat{g}(t)$ depend on time: the metric $\hat{g}(t)$ changes, and the map $\varphi_t$ also changes. The correct differentiation rule therefore has two terms.
Define $V_t: M \to TM$ by
\begin{align*}
V_t(q)=\frac{d}{dt}\varphi_t(\varphi_t^{-1}(q))
\end{align*}
for each $q \in M$. This is the Eulerian velocity field of the moving diffeomorphisms. The time-dependent identity for the [pullback](/page/Pullback) of a covariant $2$-tensor $h(t)$ is
\begin{align*}
\frac{d}{dt}\varphi_t^*h(t)=\varphi_t^*\left(\partial_t h(t)+\mathcal{L}_{V_t}h(t)\right).
\end{align*}
The term $\partial_t h(t)$ records the intrinsic time variation of the tensor, while $\mathcal{L}_{V_t}h(t)$ records the change caused by transporting the tensor along the moving diffeomorphisms.
We apply this formula with $h(t)=\hat{g}(t)$. Hence
\begin{align*}
\partial_t g(t)=\frac{d}{dt}\varphi_t^*\hat{g}(t).
\end{align*}
Applying the time-dependent pullback formula gives
\begin{align*}
\partial_t g(t)=\varphi_t^*\left(\partial_t\hat{g}(t)+\mathcal{L}_{V_t}\hat{g}(t)\right).
\end{align*}
From the defining ordinary differential equation for $\varphi_t$,
\begin{align*}
\frac{d}{dt}\varphi_t(p)=-W_t(\varphi_t(p)).
\end{align*}
Writing $q=\varphi_t(p)$ gives $p=\varphi_t^{-1}(q)$, and therefore
\begin{align*}
V_t(q)=\frac{d}{dt}\varphi_t(\varphi_t^{-1}(q))=-W_t(q).
\end{align*}
Thus $V_t=-W_t$. Since Lie differentiation is linear in the vector field,
\begin{align*}
\mathcal{L}_{V_t}\hat{g}(t)=\mathcal{L}_{-W_t}\hat{g}(t)=-\mathcal{L}_{W_t}\hat{g}(t).
\end{align*}
Substituting this into the pullback identity yields
\begin{align*}
\partial_t g(t)
=\varphi_t^*\left(\partial_t\hat{g}(t)-\mathcal{L}_{W_t}\hat{g}(t)\right).
\end{align*}
This is the step where the sign convention in the DeTurck diffeomorphism equation matters: choosing the velocity $-W_t$ is exactly what will cancel the DeTurck correction term.
[/guided]
[/step]
[step:Cancel the DeTurck Lie derivative term]
Since $\hat{g}(t)$ solves Ricci-DeTurck flow,
\begin{align*}
\partial_t\hat{g}(t)=-2\operatorname{Ric}(\hat{g}(t))+\mathcal{L}_{W_t}\hat{g}(t).
\end{align*}
Substituting this identity into the previous expression for $\partial_t g(t)$ gives
\begin{align*}
\partial_t g(t)=\varphi_t^*\left(-2\operatorname{Ric}(\hat{g}(t))+\mathcal{L}_{W_t}\hat{g}(t)-\mathcal{L}_{W_t}\hat{g}(t)\right).
\end{align*}
Cancelling the two opposite Lie derivative terms inside the pullback gives
\begin{align*}
\partial_t g(t)=-2\varphi_t^*\operatorname{Ric}(\hat{g}(t)).
\end{align*}
[/step]
[step:Use naturality of Ricci curvature under pullback]
The [Ricci curvature](/page/Ricci%20Curvature) is natural under diffeomorphism pullback: for every diffeomorphism $\psi:M\to M$ and every smooth Riemannian metric $h$ on $M$,
\begin{align*}
\psi^*\operatorname{Ric}(h)=\operatorname{Ric}(\psi^*h).
\end{align*}
Indeed, the pullback connection $\psi^*\nabla^h$ is torsion-free and compatible with $\psi^*h$, so by uniqueness of the Levi-Civita connection it equals $\nabla^{\psi^*h}$. Let $\operatorname{Rm}(h)$ denote the full curvature tensor of $\nabla^h$ as a covariant $4$-tensor. The equality of the pulled-back Levi-Civita connection with $\nabla^{\psi^*h}$ gives
\begin{align*}
\operatorname{Rm}(\psi^*h)=\psi^*\operatorname{Rm}(h).
\end{align*}
The inverse metric also pulls back naturally: the inverse of $\psi^*h$ is $\psi^*(h^{-1})$ under the induced pullback on contravariant $2$-tensors. Therefore the contraction defining Ricci curvature satisfies
\begin{align*}
\operatorname{tr}_{\psi^*h}\operatorname{Rm}(\psi^*h)
=\operatorname{tr}_{\psi^*(h^{-1})}\psi^*\operatorname{Rm}(h)
=\psi^*\left(\operatorname{tr}_{h^{-1}}\operatorname{Rm}(h)\right),
\end{align*}
which is exactly the displayed Ricci identity.
Applying this with $\psi=\varphi_t$ and $h=\hat{g}(t)$ gives
\begin{align*}
\varphi_t^*\operatorname{Ric}(\hat{g}(t))=\operatorname{Ric}(\varphi_t^*\hat{g}(t)).
\end{align*}
Since $g(t)=\varphi_t^*\hat{g}(t)$, this becomes
\begin{align*}
\varphi_t^*\operatorname{Ric}(\hat{g}(t))=\operatorname{Ric}(g(t)).
\end{align*}
Therefore
\begin{align*}
\partial_t g(t)=-2\operatorname{Ric}(g(t)).
\end{align*}
[/step]
[step:Verify the initial condition]
Since $\varphi_0=\operatorname{id}_M$, the identity diffeomorphism pulls back every covariant tensor to itself. Therefore
\begin{align*}
g(0)=\hat{g}(0).
\end{align*}
Combining this initial condition with
\begin{align*}
\partial_t g(t)=-2\operatorname{Ric}(g(t))
\end{align*}
shows that $g(t)=\varphi_t^*\hat{g}(t)$ solves Ricci flow with $g(0)=\hat{g}(0)$.
[/step]
