There is a explain on my book,as the below picture. $\rm Rm$ is curvature tensor, in fact,I don't know why the curvature tensor be denoted by $\rm Rm$, I think it's should be $R_{ijkl}$. Besides $\partial _tg_{ij}=-2R_{ij}$.
I have try to compute the below equation. But the result is $\partial_t\Gamma_{jk}^i=-g^{il}(\nabla_jR_{kl}+\nabla_kR_{jl}-\nabla_lR_{jk})$. I don't know how to make the RHS equal to $\nabla\rm Rm$,even choice normal frame ,I only have $\partial_t\Gamma_{jk}^i=-(\nabla_jR_{ki}+\nabla_kR_{ji}-\nabla_iR_{jk})$.
The below picture is from the 191th page of the paper
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This theorem is concerned with proving an estimate, and as estimates tend to go there is a lot of wiggle room in the proof, especially since we are not concerned with the exact constants involved. Thus it is commonplace not to keep track of all the little details when we are just going to estimate them all (usually with Cauchy-Schwarz) and lump them together anyway.
In this case, the exact expression is $\partial_t \Gamma^i_{jk} = -g^{il}\left(\nabla_j R_{kl} + \nabla_k R_{jl} - \nabla_l R_{jk}\right)$; but since we are just going to be taking norms later, all that matters is that it's first order in $\nabla Rm$, i.e. a linear combination of some components of $\nabla Rm$.
Here's how we use this to get (1.4.2); hopefully this will help you understand the idea: we take the time derivative of $\nabla Rm$ to get $$\partial_t \nabla Rm = \partial_t (\partial Rm + \Gamma * Rm) = \partial \partial_t Rm + \Gamma * \partial_t Rm + \partial_t \Gamma * Rm.$$
(Note the complete lack of care about how the indices go together here - we really don't care.) Now recognise $\partial + \Gamma$ as a covariant derivative (OK, strictly this part should be done more precisely, but it works out) to get $$\partial_t \nabla Rm = \nabla(\partial_t Rm) + \partial_t \Gamma * Rm.$$
Since we know $\partial_t Rm = \Delta Rm + Rm * Rm$ and $\partial_t \Gamma =^* \nabla Rm$, we get
$$\partial_t \nabla Rm = \nabla \Delta Rm + \nabla (Rm * Rm) + \nabla Rm * Rm.$$
The product rule tells us that $\nabla (Rm * Rm) = \nabla Rm * Rm$, and commuting derivatives with the curvature tensor tells us $\nabla \Delta Rm = \Delta \nabla Rm + \nabla Rm * Rm$, and thus we can conclude (rolling all the $\nabla Rm * Rm$ terms into one - if we were keeping track of constants we'd add them)
$$ \partial_t \nabla Rm = \Delta\nabla Rm+ \nabla Rm * Rm.$$
Personally I think it's a little sloppy to write the equation $\partial_t \Gamma\ ``=" \nabla Rm$ as these authors have. It's clear in context what they mean, but I think it's better to always use the $*$ notation or similar, especially since it's important that the $\Delta \nabla Rm$ term is actually exact. I can understand the choice, though - you either have to invent a new notation for the unary case or write something silly like $\nabla Rm * 1$.