I'm trying to understand this lemma on "Hamilton's Ricci Flow", by Ben Chow.

but I can't see how he got to the equality outlined in red (the author mentions the $\cdot$ stands for dot product or standard multiplication). I've done some work already but didn't quite get what I want. Now, it's easy to show that:
$$ \frac{\partial}{\partial t} g^{i j}=-g^{i k} g^{j \ell} h_{k \ell} $$
where I'm assuming that $$ \frac{\partial}{\partial t} g_{i j}=h_{i j} $$ where $h$ is some symmetric $2$-tensor. Now, if $f$ is an arbitrary smooth function on $M$, the product rule gives us:
$$ \begin{aligned} \frac{\partial}{\partial t}(\Delta f) &=\frac{\partial}{\partial t}\left[g^{i j}\left(\partial_{i} \partial_{j}-\Gamma_{i j}^{k} \partial_{k}\right) f\right] \\ &=\left(\frac{\partial}{\partial t} g^{i j}\right) \nabla_{i} \nabla_{j} f-g^{i j}\left(\frac{\partial}{\partial t} \Gamma_{i j}^{k}\right) \nabla_{k} f+\Delta\left(\frac{\partial}{\partial t} f\right) \end{aligned} $$
and we could replace $\frac{\partial}{\partial t} g^{i j}$ with the expression found earlier, but I can't see how that'd give us the part outlined in red. Comparing both expressions, I guess my doubt is how to show the following equality:
$$ -\frac{\partial}{\partial t} g_{i j} \cdot \nabla_{i} \nabla_{j} = \left(\frac{\partial}{\partial t} g^{i j}\right) \nabla_{i} \nabla_{j} +\Delta\left(\frac{\partial}{\partial t} \right) $$
which is where I really got stuck. I'm grateful for any help.
I think you're basically there, once you're aware of a couple assumptions/conventions: