[proofplan]
We differentiate the coordinate formula for the Levi-Civita connection of $g(t)$. The only delicate point is the derivative of the inverse metric; using the differentiated identity $g^{ka}g_{a\ell}=\delta^k_\ell$, that term becomes exactly the Christoffel-symbol correction term needed to convert partial derivatives of $v$ into covariant derivatives. Finally, the Ricci flow formula follows by substituting $v_{ij}=-2\operatorname{Ric}_{ij}$.
[/proofplan]
[step:Differentiate the inverse metric identity]
Fix $t \in I$ and work in the coordinate chart $(U,x)$. Let $G(t)$ denote the matrix-valued function $G(t)=(g_{ij}(t))$, and let $G(t)^{-1}=(g^{ij}(t))$. Let $\delta^k_\ell$ denote the Kronecker delta, defined by $\delta^k_\ell=1$ when $k=\ell$ and $\delta^k_\ell=0$ when $k\ne \ell$. Since
\begin{align*}
\sum_{a=1}^n g^{ka}g_{a\ell}=\delta^k_\ell,
\end{align*}
differentiating with respect to $t$ gives
\begin{align*}
\sum_{a=1}^n (\partial_t g^{ka})g_{a\ell}
+
\sum_{a=1}^n g^{ka}v_{a\ell}
=
0.
\end{align*}
Multiplying by $g^{\ell b}$ and summing over $\ell$ yields
\begin{align*}
\partial_t g^{kb}
=
-\sum_{a,\ell=1}^n g^{ka}g^{b\ell}v_{a\ell}.
\end{align*}
Equivalently,
\begin{align*}
\partial_t g^{k\ell}
=
-g^{ka}g^{\ell b}v_{ab}.
\end{align*}
[guided]
The inverse metric also depends on $t$, so differentiating Christoffel symbols requires differentiating $g^{k\ell}$ as well as $g_{ij}$. The relation defining the inverse matrix is
\begin{align*}
\sum_{a=1}^n g^{ka}g_{a\ell}=\delta^k_\ell.
\end{align*}
Here $\delta^k_\ell$ denotes the Kronecker delta, defined by $\delta^k_\ell=1$ when $k=\ell$ and $\delta^k_\ell=0$ when $k\ne \ell$. It is constant in $t$, while $\partial_t g_{a\ell}=v_{a\ell}$ by definition of $v$. Differentiating the identity gives
\begin{align*}
\sum_{a=1}^n (\partial_t g^{ka})g_{a\ell}
+
\sum_{a=1}^n g^{ka}v_{a\ell}
=
0.
\end{align*}
To solve for $\partial_t g^{kb}$, multiply by $g^{\ell b}$ and sum in $\ell$. Since $\sum_{\ell=1}^n g_{a\ell}g^{\ell b}=\delta_a^b$, the first term reduces to $\partial_t g^{kb}$, and we obtain
\begin{align*}
\partial_t g^{kb}
=
-\sum_{a,\ell=1}^n g^{ka}g^{b\ell}v_{a\ell}.
\end{align*}
Renaming the free index $b$ as $\ell$ gives
\begin{align*}
\partial_t g^{k\ell}
=
-g^{ka}g^{\ell b}v_{ab}.
\end{align*}
This identity is the source of the cancellation in the variation formula.
[/guided]
[/step]
[step:Differentiate the coordinate formula for the Levi-Civita connection]
For each $t \in I$, the Christoffel symbols of $\nabla(t)$ satisfy
\begin{align*}
\Gamma_{ij}^k
=
\frac{1}{2}g^{k\ell}
\left(
\partial_{x_i}g_{j\ell}
+
\partial_{x_j}g_{i\ell}
-
\partial_{x_\ell}g_{ij}
\right).
\end{align*}
Define the coordinate expression
\begin{align*}
S_{ij\ell}
:=
\partial_{x_i}g_{j\ell}
+
\partial_{x_j}g_{i\ell}
-
\partial_{x_\ell}g_{ij}.
\end{align*}
Then
\begin{align*}
S_{ij\ell}=2g_{\ell m}\Gamma_{ij}^m.
\end{align*}
Because $g(t)$ is a smooth one-parameter family of metrics, the coordinate functions $(x,t)\mapsto g_{ij}(x,t)$ are smooth on $U\times I$, so the spatial and time derivatives commute. Differentiating $\Gamma_{ij}^k=\frac{1}{2}g^{k\ell}S_{ij\ell}$ with respect to $t$ gives
\begin{align*}
\partial_t\Gamma_{ij}^k=\frac{1}{2}(\partial_t g^{k\ell})S_{ij\ell}+\frac{1}{2}g^{k\ell}\left(\partial_{x_i}v_{j\ell}+\partial_{x_j}v_{i\ell}-\partial_{x_\ell}v_{ij}\right).
\end{align*}
Using $\partial_t g^{k\ell}=-g^{ka}g^{\ell b}v_{ab}$ and $S_{ij\ell}=2g_{\ell m}\Gamma_{ij}^m$, this becomes
\begin{align*}
\partial_t\Gamma_{ij}^k=\frac{1}{2}g^{k\ell}\left(\partial_{x_i}v_{j\ell}+\partial_{x_j}v_{i\ell}-\partial_{x_\ell}v_{ij}\right)-g^{ka}v_{am}\Gamma_{ij}^m.
\end{align*}
[/step]
[step:Rewrite the differentiated formula using covariant derivatives of $v$]
For a covariant $2$-tensor $v$, its covariant derivative with respect to $\nabla(t)$ is given in coordinates by
\begin{align*}
\nabla_i v_{j\ell}
=
\partial_{x_i}v_{j\ell}
-
\Gamma_{ij}^m v_{m\ell}
-
\Gamma_{i\ell}^m v_{jm}.
\end{align*}
Therefore, expanding each of the three covariant derivative terms gives
\begin{align*}
\nabla_i v_{j\ell}+\nabla_j v_{i\ell}-\nabla_\ell v_{ij}=\partial_{x_i}v_{j\ell}+\partial_{x_j}v_{i\ell}-\partial_{x_\ell}v_{ij}-\Gamma_{ij}^m v_{m\ell}-\Gamma_{i\ell}^m v_{jm}-\Gamma_{ji}^m v_{m\ell}-\Gamma_{j\ell}^m v_{im}+\Gamma_{\ell i}^m v_{mj}+\Gamma_{\ell j}^m v_{im}.
\end{align*}
Since the Levi-Civita connection is torsion-free, $\Gamma_{ij}^m=\Gamma_{ji}^m$, $\Gamma_{i\ell}^m=\Gamma_{\ell i}^m$, and $\Gamma_{j\ell}^m=\Gamma_{\ell j}^m$. Since $v$ is symmetric, $v_{jm}=v_{mj}$. Thus the middle terms cancel except for the two identical $\Gamma_{ij}^m v_{m\ell}$ terms, and
\begin{align*}
\nabla_i v_{j\ell}+\nabla_j v_{i\ell}-\nabla_\ell v_{ij}=\partial_{x_i}v_{j\ell}+\partial_{x_j}v_{i\ell}-\partial_{x_\ell}v_{ij}-2\Gamma_{ij}^m v_{m\ell}.
\end{align*}
Multiplying by $\frac{1}{2}g^{k\ell}$ gives
\begin{align*}
\frac{1}{2}g^{k\ell}\left(\nabla_i v_{j\ell}+\nabla_j v_{i\ell}-\nabla_\ell v_{ij}\right)=\frac{1}{2}g^{k\ell}\left(\partial_{x_i}v_{j\ell}+\partial_{x_j}v_{i\ell}-\partial_{x_\ell}v_{ij}\right)-g^{k\ell}\Gamma_{ij}^m v_{m\ell}.
\end{align*}
Renaming the summed index $\ell$ as $a$ in the last term, this is exactly the expression obtained for $\partial_t\Gamma_{ij}^k$. Hence
\begin{align*}
\partial_t\Gamma_{ij}^k
=
\frac{1}{2}g^{k\ell}
\left(
\nabla_i v_{j\ell}
+
\nabla_j v_{i\ell}
-
\nabla_\ell v_{ij}
\right).
\end{align*}
[/step]
[step:Insert the Ricci flow variation tensor]
Assume now that $g(t)$ evolves by Ricci flow, so
\begin{align*}
v_{ij}=\partial_t g_{ij}=-2\operatorname{Ric}_{ij}.
\end{align*}
Substituting this into the already proved variation formula gives
\begin{align*}
\partial_t\Gamma_{ij}^k=\frac{1}{2}g^{k\ell}\left(\nabla_i(-2\operatorname{Ric}_{j\ell})+\nabla_j(-2\operatorname{Ric}_{i\ell})-\nabla_\ell(-2\operatorname{Ric}_{ij})\right).
\end{align*}
By linearity of the covariant derivative in the tensor slot, this simplifies to
\begin{align*}
\partial_t\Gamma_{ij}^k=-g^{k\ell}\left(\nabla_i\operatorname{Ric}_{j\ell}+\nabla_j\operatorname{Ric}_{i\ell}-\nabla_\ell\operatorname{Ric}_{ij}\right).
\end{align*}
This is the stated Ricci flow specialization.
[/step]
