Comment: the following is a somewhat convoluted way of deriving the Euler-Lagrange equation using Clairaut's theorem for the volume functional and some standard, albeit not simpler, variation formulas (all is $C^{\infty}$ in the following). Let $g_{0}$ be a Riemannian metric and let $v$ be a symmetric $2$-tensor. Let $g_{0,s}$ satisfy $\frac{\partial}{\partial s}|_{s=0} g_{0,s}=v$ and $g_{0,0}=g_{0}$. With $g_{0,s}$ as initial data, let $g_{t,s}$, $t\in\lbrack0,\varepsilon)$, solve the Ricci-Hamilton-DeTurck flow $\frac{\partial}{\partial t}g_{t,s}=-2\operatorname{Ric}{}_{g_{t,s} }+\mathcal{L}_{W_{t,s}}g_{t,s}$, where $W_{t,s}=\operatorname{tr}^{1,2} {}_{g_{t,s}}(\nabla_{g_{t,s}}-\nabla_{g_{t,0}})$. Note that $\frac{\partial }{\partial t}g_{t,0}=-2\operatorname{Ric}{}_{g_{t,0}}$. Let $v_{t,s} =\frac{\partial}{\partial s}g_{t,s}$. We have $\frac{\partial^{2}}{\partial s\partial t}|_{s=0}g_{t,s}=\Delta_{L}v_{t,0}$, which equals $\frac {\partial^{2}}{\partial t\partial s}|_{s=0}g_{t,s}=\frac{\partial}{\partial t}v_{t,0}$ (Lichnerowicz Laplacian heat equation). We compute $\frac{\partial }{\partial t}\operatorname{Vol}(g_{t,s})=\frac{1}{2}\int\operatorname{tr} {}_{g_{t,s}}(\frac{\partial}{\partial t}g_{t,s})d\mu_{g_{t,s}}=-\int R_{g_{t,s}}d\mu_{g_{t,s}}$ since $\int\operatorname{tr}{}_{g_{t,s} }(\mathcal{L}_{W_{t,s}}g_{t,s})d\mu_{g_{t,s}}=0$ by the divergence theorem. Now \begin{align*} \frac{\partial}{\partial s}|_{s=0}\int R_{g_{t,s}}d\mu_{g_{t,s}} & =-\frac{\partial^{2}}{\partial s\partial t}|_{s=0}\operatorname{Vol} (g_{t,s})=-\frac{\partial^{2}}{\partial t\partial s}|_{s=0}\operatorname{Vol} (g_{t,s})\\ & =-\frac{1}{2}\frac{\partial}{\partial t}\int\operatorname{tr}{}_{g_{t,0} }(v_{t,0})d\mu_{g_{t,0}}\\ & =\int\langle-\operatorname{Ric}{}_{g_{t,0}}+\frac{g_{t,0}}{2}R_{g_{t,0} },v_{t,0}\rangle_{g_{t,0}}d\mu_{g_{g_{t,0}}} \end{align*} since $\int\operatorname{tr}{}_{g_{t,0}}(\frac{\partial}{\partial t} v_{t,0})d\mu_{g_{t,0}}=\int\operatorname{tr}{}_{g_{t,0}}(\Delta_{L} v_{t,0})d\mu_{g_{t,0}}=\int\Delta_{g_{t,0}}(\operatorname{tr}{}_{g_{t,0} }(v_{t,0}))d\mu_{g_{t,0}}=0$. Finally, take $t=0$.

Comment: the following is a somewhat convoluted way of deriving the Euler-Lagrange equation using Clairaut's theorem for the volume functional and some standard, albeit not simpler, variation formulas (all is $C^{\infty}$ in the following). Let $g_{0}$ be a Riemannian metric and let $v$ be a symmetric $2$-tensor. Let $g_{0,s}$ satisfy $\frac{\partial}{\partial s}|_{s=0} g_{0,s}=v$ and $g_{0,0}=g_{0}$. With $g_{0,s}$ as initial data, let $g_{t,s}$, $t\in\lbrack0,\varepsilon)$, solve the Ricci-Hamilton-DeTurck flow $\frac{\partial}{\partial t}g_{t,s}=-2\operatorname{Ric}{}_{g_{t,s} }+\mathcal{L}_{W_{t,s}}g_{t,s}$, where $W_{t,s}=\operatorname{tr}^{1,2} {}_{g_{t,s}}(\nabla_{g_{t,s}}-\nabla_{g_{t,0}})$. Note that $\frac{\partial }{\partial t}g_{t,0}=-2\operatorname{Ric}{}_{g_{t,0}}$. Let $v_{t,s} =\frac{\partial}{\partial s}g_{t,s}$. We have $\frac{\partial^{2}}{\partial s\partial t}|_{s=0}g_{t,s}=\Delta_{L}v_{t,0}$, which equals $\frac {\partial^{2}}{\partial t\partial s}|_{s=0}g_{t,s}=\frac{\partial}{\partial t}v_{t,0}$ (Lichnerowicz Laplacian heat equation). We compute $\frac{\partial }{\partial t}\operatorname{Vol}(g_{t,s})=\frac{1}{2}\int\operatorname{tr} {}_{g_{t,s}}(\frac{\partial}{\partial t}g_{t,s})d\mu_{g_{t,s}}=-\int R_{g_{t,s}}d\mu_{g_{t,s}}$ since $\int\operatorname{tr}{}_{g_{t,s} }(\mathcal{L}_{W_{t,s}}g_{t,s})d\mu_{g_{t,s}}=0$ by the divergence theorem. Now \begin{align*} \frac{\partial}{\partial s}|_{s=0}\int R_{g_{t,s}}d\mu_{g_{t,s}} & =-\frac{\partial^{2}}{\partial s\partial t}|_{s=0}\operatorname{Vol} (g_{t,s})=-\frac{\partial^{2}}{\partial t\partial s}|_{s=0}\operatorname{Vol} (g_{t,s})\\ & =-\frac{1}{2}\frac{\partial}{\partial t}\int\operatorname{tr}{}_{g_{t,0} }(v_{t,0})d\mu_{g_{t,0}}\\ & =\int\langle-\operatorname{Ric}{}_{g_{t,0}}+\frac{g_{t,0}}{2}R_{g_{t,0} },v_{t,0}\rangle_{g_{t,0}}d\mu_{g_{g_{t,0}}} \end{align*} since $\int\operatorname{tr}{}_{g_{t,0}}(\frac{\partial}{\partial t} v_{t,0})d\mu_{g_{t,0}}=\int\operatorname{tr}{}_{g_{t,0}}(\Delta_{L} v_{t,0})d\mu_{g_{t,0}}=\int\Delta_{g_{t,0}}(\operatorname{tr}{}_{g_{t,0} }(v_{t,0}))d\mu_{g_{t,0}}=0$. Finally, take $t=0$.

December 18, 2013. The notion of volume, in various guises, occurs throughout the study of Ricci flow, especially in Perelman's work. Now, per unit increase in scale $t$, the volume form of a metric changes with velocity $\frac {\partial}{\partial t}d\mu=-Rd\mu$. By Clairaut's theorem, the variation of $-Rd\mu$ is equal to the change per unit increase in scale of the variation of the volume form, i.e., $$ \frac{\partial}{\partial t}(\frac{\operatorname{tr}_{g}v}{2}d\mu_{g})=\left( \langle\operatorname{Ric}-\frac{1}{2}Rg,v\rangle+\operatorname{div} (\frac{\nabla\operatorname{tr}v}{2})\right) d\mu. $$

In the $f$-warped or entropy version of this, we have $\frac{\partial }{\partial t}(fe^{-f}d\mu)=(-R-\Delta f)e^{-f}d\mu$ under $\frac{\partial }{\partial t}g=-2(\operatorname{Ric}+\nabla^{2}f)$ and $\frac{\partial f}{\partial t}=-R-\Delta f$. Integrating this yields that $\mathcal{N} \doteqdot\int_{\mathcal{M}}fe^{-f}d\mu$ satisfies $-\frac{d\mathcal{N}} {dt}=\mathcal{F}\doteqdot\int(R+|\nabla f|^{2})e^{-f}d\mu$. If $\frac {\partial}{\partial s}g=v$ and $\frac{\partial f}{\partial s}=\frac {\operatorname{tr}_{g}v}{2}$, then the variation of the energy integrand is \begin{align*} & \frac{\partial}{\partial s}((-R-\Delta f)e^{-f}d\mu)\\ & =\left( (-L(v,\nabla f)+2\langle\operatorname{Ric}+\nabla^{2} f,v\rangle)e^{-f}+\operatorname{div}(e^{-f}\{\frac{\nabla\operatorname{tr} v}{2}-v(\nabla f)\})\right) d\mu\doteqdot A, \end{align*} where $L(v,X)=\operatorname{div}^{2}v+\langle\operatorname{Ric},v\rangle -2\langle\operatorname{div}v,X\rangle+v(X,X)$ is the linear trace Harnack quadratic. On the other hand, $\frac{\partial}{\partial s}(fe^{-f}d\mu )=\frac{\operatorname{tr}_{g}v}{2}e^{-f}d\mu$. So Perelman's version is $\frac{\partial}{\partial t}(\frac{\operatorname{tr}_{g}v}{2}e^{-f}d\mu )=\frac{\partial^{2}}{\partial s\partial t}(fe^{-f}d\mu)=A$, using Clairaut's theorem. Note that integration by parts gives $\int L(v,\nabla f)e^{-f} d\mu=\int\langle\operatorname{Ric}+\nabla^{2}f,v\rangle e^{-f}d\mu$, from which one obtains Perelman's energy variation formula.

In Section 6.2 of arXiv:0211159 Perelman argues that the $\mathcal{W}$-entropy (i.e., $\mathcal{F}$ with scaling) integrand is a warped scalar curvature. So, without scaling (i.e., $\tau$), we would have the correspondences $\mathcal{N}\sim\operatorname{Vol}$ and $\mathcal{F}\sim\int Rd\mu$, which is also clear from taking $f=\operatorname{const}\neq0$ as a special case. However, in 6.2, Perelman's volume is essentially $\int e^{-f}d\mu$, which is constant under the above variations.