In page 2 of An optimal volume growth estimate for noncollapsed steady gradient Ricci solitons, the authors say the blow result.
Assum $(M,g)$ is a Riemannian manifold, $f:M\rightarrow \mathbb R$ is a function, $k$ is a constant, $\Phi_t:M\rightarrow M$ are diffeomorphisms. If \begin{align} & \partial_t \Phi_t = \frac{1}{1-kt}\nabla f \circ \Phi_t \\ & \Phi_0= id \\ & g_t = (1-kt)\Phi_t^* g \end{align} and $g$ satisfy $$ Ric +\nabla^2 f = \frac{k}{2}g $$ then $g_t$ moves by Ricci flow.
I want to verify it. But, I don't know how to use $$ \partial_t \Phi_t = \frac{1}{1-kt}\nabla f \circ \Phi_t $$ to calculate $$ \partial_t [\Phi_t^* g]. $$
After Dider's hint, (although I don't know why $\partial_t$ and $d$ commute)I get $$ \partial_t[\Phi_t^* g]= 2g_{\Phi_t}(d\partial_t\Phi_t\cdot,d\Phi_t\cdot) \\ = 2g_{\Phi_t}(d[\frac{1}{1-kt}\nabla f\circ \Phi_t]\cdot,d\Phi_t\cdot) \\ = 2\frac{1}{1-kt}d(\nabla f) g_{\Phi_t}(d\Phi_t \cdot,d\Phi_t\cdot) \\ =2\frac{1}{1-kt}d(\nabla f) \Phi_t^* g $$ Therefore $$ \partial_t g_t =\Phi_t^*(-kg+ 2d(\nabla f)g) $$ I guess $d(\nabla f)g=\nabla^2 f$, but don't know why. In fact, I don't know what is the differentiation of a vector field.
Besides, why the $\Phi_t$ of $g_{\Phi_t}$ needn't to be $\partial_t$ ? (If need, I don't know how to calculate it.)