[proofplan] The proof is local, because equality of smooth top-degree forms can be checked in oriented coordinate charts. In such a chart, the Riemannian volume form is expressed as a positive density multiplying the coordinate orientation form. Differentiating that density in time and using Jacobi's formula for the derivative of the determinant gives the factor involving the metric trace. The Ricci flow case follows by substituting the Ricci tensor variation and using the definition of scalar curvature. [/proofplan]

[step:Write the volume form in a positively oriented coordinate chart] Fix $t_0 \in I$ and a point $p \in M$. Let $(U,\varphi)$ be a positively oriented smooth coordinate chart with $p \in U$, where $\varphi: U \to \varphi(U) \subseteq \mathbb{R}^n$ is the coordinate map and $\varphi(q)=(x_1(q),\dots,x_n(q))$. For $q \in U$ and $t \in I$, define the coordinate matrices \begin{align*} G(q,t) := \bigl(g_{ij}(q,t)\bigr)_{i,j=1}^n. \end{align*} Define also \begin{align*} V(q,t) := \bigl(v_{ij}(q,t)\bigr)_{i,j=1}^n. \end{align*} Here the metric coefficients are \begin{align*} g_{ij}(q,t) := g(t)_q(\partial_{x_i}|_q,\partial_{x_j}|_q). \end{align*} The variation coefficients are \begin{align*} v_{ij}(q,t) := v(t)_q(\partial_{x_i}|_q,\partial_{x_j}|_q). \end{align*} Since $g(t)$ is a Riemannian metric, $G(q,t)$ is symmetric positive definite, so $\det G(q,t)>0$. By the definition of the Riemannian volume form, in this chart, \begin{align*} d\mu_{g(t)} = \sqrt{\det G(q,t)}\,dx_1 \wedge \cdots \wedge dx_n. \end{align*} Thus it suffices to differentiate the scalar coefficient $\sqrt{\det G(q,t)}$. [/step]

[step:Differentiate the determinant coefficient]Fix $q \in U$ and $t \in I$. We use Jacobi's formula for the derivative of the determinant for the smooth curve of invertible matrices $t \mapsto G(q,t)$: \begin{align*} \partial_t \det G(q,t) = \det G(q,t)\,\operatorname{tr}\bigl(G(q,t)^{-1}V(q,t)\bigr). \end{align*} Writing $G(q,t)^{-1}=(g^{ij}(q,t))_{i,j=1}^n$, this becomes \begin{align*} \partial_t \det G(q,t) = \det G(q,t)\sum_{i,j=1}^n g^{ij}(q,t)v_{ij}(q,t). \end{align*} By the coordinate formula for the metric trace, \begin{align*} \partial_t \det G(q,t) = \det G(q,t)\operatorname{tr}_{g(t)}(v(t))(q). \end{align*} Since $\det G(q,t)>0$, the ordinary chain rule applied to $r \mapsto \sqrt r$ gives \begin{align*} \partial_t\sqrt{\det G(q,t)} = \frac{1}{2}(\det G(q,t))^{-1/2}\partial_t\det G(q,t). \end{align*} Substituting the determinant derivative gives \begin{align*} \partial_t\sqrt{\det G(q,t)} = \frac{1}{2}\operatorname{tr}_{g(t)}(v(t))(q)\sqrt{\det G(q,t)}. \end{align*}[/step]

[guided]The only real calculation in the proof is the derivative of the determinant. We are working in the coordinate chart $(U,\varphi)$ fixed above, and for $q \in U$ and $t \in I$ the matrices are $G(q,t)=(g_{ij}(q,t))_{i,j=1}^n$ and $V(q,t)=(v_{ij}(q,t))_{i,j=1}^n$. Fix $q \in U$ and consider the smooth matrix-valued curve $A: I \to \operatorname{Sym}^+_n$ defined by $A(t)=G(q,t)$, where $\operatorname{Sym}^+_n$ denotes the set of positive definite symmetric real $n \times n$ matrices. Its derivative is the matrix-valued curve $A': I \to \operatorname{Sym}_n$ defined by $A'(t)=V(q,t)$, because $v(t)=\partial_t g(t)$. For an invertible matrix curve $A(t)$, Jacobi's formula for the derivative of the determinant says \begin{align*} \frac{d}{dt}\det A(t)=\det A(t)\operatorname{tr}\bigl(A(t)^{-1}A'(t)\bigr). \end{align*} Applying this to $A(t)=G(q,t)$ gives \begin{align*} \partial_t \det G(q,t) = \det G(q,t)\operatorname{tr}\bigl(G(q,t)^{-1}V(q,t)\bigr). \end{align*} If $G(q,t)^{-1}=(g^{ij}(q,t))_{i,j=1}^n$, then the trace of the matrix product is \begin{align*} \operatorname{tr}\bigl(G(q,t)^{-1}V(q,t)\bigr) = \sum_{i,j=1}^n g^{ij}(q,t)v_{ij}(q,t). \end{align*} This is exactly the coordinate expression for the metric trace of $v(t)$ with respect to $g(t)$: \begin{align*} \operatorname{tr}_{g(t)}(v(t))(q) = \sum_{i,j=1}^n g^{ij}(q,t)v_{ij}(q,t). \end{align*} Therefore \begin{align*} \partial_t \det G(q,t) = \det G(q,t)\operatorname{tr}_{g(t)}(v(t))(q). \end{align*} Now we pass from the determinant to the square root of the determinant. The positivity of $G(q,t)$ implies $\det G(q,t)>0$, so the function $r \mapsto \sqrt r$ is smooth at $r=\det G(q,t)$. The ordinary chain rule gives \begin{align*} \partial_t\sqrt{\det G(q,t)} = \frac{1}{2}(\det G(q,t))^{-1/2}\partial_t\det G(q,t). \end{align*} Substituting the determinant derivative gives \begin{align*} \partial_t\sqrt{\det G(q,t)} = \frac{1}{2}(\det G(q,t))^{-1/2}\det G(q,t)\operatorname{tr}_{g(t)}(v(t))(q). \end{align*} Cancelling the power of $\det G(q,t)$ yields \begin{align*} \partial_t\sqrt{\det G(q,t)} = \frac{1}{2}\operatorname{tr}_{g(t)}(v(t))(q)\sqrt{\det G(q,t)}. \end{align*} This is the origin of the factor $\frac{1}{2}$ in the volume evolution formula.[/guided]

[step:Convert the coefficient identity into an identity of volume forms] Using the coordinate expression from the first step and the coefficient derivative from the second step, we compute on $U$. The coordinate form gives \begin{align*} \partial_t d\mu_{g(t)} = \partial_t\bigl(\sqrt{\det G(q,t)}\,dx_1 \wedge \cdots \wedge dx_n\bigr). \end{align*} The coordinate one-forms are independent of $t$, so \begin{align*} \partial_t d\mu_{g(t)} = \bigl(\partial_t\sqrt{\det G(q,t)}\bigr)\,dx_1 \wedge \cdots \wedge dx_n. \end{align*} Substituting the coefficient derivative gives \begin{align*} \partial_t d\mu_{g(t)} = \frac{1}{2}\operatorname{tr}_{g(t)}(v(t))(q)\sqrt{\det G(q,t)}\,dx_1 \wedge \cdots \wedge dx_n. \end{align*} Using again the coordinate expression for $d\mu_{g(t)}$, this becomes \begin{align*} \partial_t d\mu_{g(t)} = \frac{1}{2}\operatorname{tr}_{g(t)}(v(t))\,d\mu_{g(t)}. \end{align*} Since $p \in M$ and the positively oriented coordinate chart around $p$ were arbitrary, the identity holds globally on $M$: \begin{align*} \partial_t d\mu_{g(t)} = \frac{1}{2}\operatorname{tr}_{g(t)}(v(t))\,d\mu_{g(t)}. \end{align*} [/step]

[step:Specialize the variation formula to Ricci flow] Assume now that $g(t)$ satisfies the Ricci flow equation involving the Ricci tensor: \begin{align*} \partial_t g(t) = -2\operatorname{Ric}_{g(t)}. \end{align*} By definition of $v(t)$, this means \begin{align*} v(t) = -2\operatorname{Ric}_{g(t)}. \end{align*} Taking the trace with respect to $g(t)$ gives \begin{align*} \operatorname{tr}_{g(t)}(v(t)) = \operatorname{tr}_{g(t)}\bigl(-2\operatorname{Ric}_{g(t)}\bigr). \end{align*} By linearity of the trace, \begin{align*} \operatorname{tr}_{g(t)}(v(t)) = -2\operatorname{tr}_{g(t)}\operatorname{Ric}_{g(t)}. \end{align*} By the definition of scalar curvature, \begin{align*} \operatorname{tr}_{g(t)}(v(t)) = -2S_{g(t)}. \end{align*} Substituting this into the general volume variation formula yields \begin{align*} \partial_t d\mu_{g(t)} = \frac{1}{2}\operatorname{tr}_{g(t)}(v(t))\,d\mu_{g(t)}. \end{align*} Therefore \begin{align*} \partial_t d\mu_{g(t)} = \frac{1}{2}(-2S_{g(t)})\,d\mu_{g(t)}. \end{align*} Hence \begin{align*} \partial_t d\mu_{g(t)} = -S_{g(t)}\,d\mu_{g(t)}. \end{align*} This is the asserted Ricci flow evolution equation for the Riemannian volume form. [/step]