The Ricci flow and $\frac{\partial}{\partial t}g_{ij}=-2(R_{ij}+\nabla_i \nabla_j f)$ are equivalent up to diffeomorphism

Suppose $M$ is a Riemannian manifold. Consider flow $\frac{\partial}{\partial t}g_{ij}=-2(R_{ij}+\nabla_i \nabla_j f)$, where $f$ is a time-dependent function. I would like to prove that flows of this form are equivalent, up to diffeomorphism, to the Ricci flow $\frac{\partial}{\partial t}g_{ij}=-2R_{ij}$, that is:

By defining a 1-parameter family of diffeomorphism $\Psi(t):M\to M$ by

$$\frac{d}{dt}\Psi(t)=\nabla_{g(t)}f(t),$$ $$\Psi(0)=id_M$$

I want to show that $\bar{g}(t):=\Psi(t)^*g(t)$ satisfy $$\frac{\partial}{\partial t}\bar{g}_{ij}=-2 \bar{R}_{ij}.$$

My problem is that I don't know how to calculate $\frac{\partial}{\partial t}\Psi(t)^*g(t)$. I know that $\frac{\partial}{\partial t}\Psi(t)^* \alpha=\mathcal{L}_{\nabla f} \alpha$, where $\mathcal{L}$ is Lie derivative and $\alpha$ is a time-independent object, but I faced problem when $\alpha$ is a time-dependent object.

Can someone point me in the right direction? Thanks in advance for your time.

You can get the correct answer by applying the formal Leibniz/product rule to the expression:

\begin{eqnarray*} \frac{\partial}{\partial t}\left(\Psi\left(t\right)^{*}g\left(t\right))\right) & = & \frac{\partial\Psi\left(t\right)^{*}}{\partial t}g\left(t\right)+\Psi\left(t\right)^{*}\frac{\partial g\left(t\right)}{\partial t}\\ & = & \Psi^{*}\left(\mathcal{L}_{\nabla f}g\right)-2\Psi^{*}\left(\textrm{Rc}+\nabla\nabla f\right)\\ & = & \Psi^{*}\left(-2\textrm{Rc}\right). \end{eqnarray*}

This calculation is written out in slightly more detail in Chapter 9.4 of http://maths-people.anu.edu.au/~andrews/book.pdf.

For a more rigorous proof, start by defining $f(u,v) = \Psi(u)^* g(v),$ so that we are trying to compute $\frac{d}{dt}f(t,t).$ Since everything here is smooth, we can apply the multivariable chain rule to rewrite this as

$$ \begin{eqnarray*} \frac{d}{dt}f(t,t) & = & \left(\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}\right)_{u=t,\;v=t} \\ & = & \left( \frac{\partial}{\partial u}\left(\Psi(u)^{*}g(v)\right) + \frac{\partial}{\partial v}\left(\Psi(u)^{*}g(v)\right) \right)_{u=t,\;v=t} \end{eqnarray*}. $$

The first term here is $\Psi(u)^*\left(\mathcal L_{\nabla f}g(v)\right)$ by the definition of the Lie derivative, and the second term is $\Psi(u)^* \frac{\partial}{\partial t}g(v)$ because $\Psi(u)^*$ is pointwise linear. Then just make the substitutions $u = t, v = t$ to obtain the result I claimed earlier.