I'm studying by myself Mean Curvature Flow and I'm trying understand the definition of $\ast$-operation given by Richard Hamilton in the beginning of the section $13$ (page $40$) of his paper "Three-manifolds with positive Ricci curvature":

"If $A$ and $B$ are two tensors we write $A \ast B$ for any linear combination of tensors formed by contraction on $A_{i \cdots j}B_{k \cdots l}$ using the $g^{ik}$".

I'm trying understand this definition in order to understand how Huisken obtained the evolution equation of Chistoffel symbols in the beginning of the section $7$ (final of the page $19$) on his paper "Flow by mean curvature of convex surfaces into spheres":

$$\frac{\partial \Gamma^i_{jk}}{\partial t} = \nabla A \ast A.$$

I found a master thesis which do the computation of this evolution equation as follows

I'm stuck in how to develop these two equalities where appears $\ast$ operation. I will put what I did in order to understand $\ast$ operation and develop these two equalities.

$\textbf{My attempt:}$

I don't know if I was clear the first time, but my doubt is what is the $\ast$-operation, because by the Hamilton's explanation on his paper lead me to think that $\ast$-operation is such that all terms of the linear combination must have a contraction of at least one of the tensors involved on $\ast$-operation, so $T \ast S$ would be one of the following forms: $\sum_{p = 1}^{p = r} \sum_{q = 1}^{q = s} \text{tr}_g \ T \ S_{i_pj_q}$, $\sum_{p = 1}^{p = r} \sum_{q = 1}^{q = s} T_{i_pj_q} \ \text{tr}_g \ S$ or $\text{tr}_g \ T \text{tr}_g \ S$, where $1 \leq i_1 < i_2 < \cdots < i_p \leq n$, $1 \leq j_1 < j_2 < \cdots < j_q \leq n$ and $p,q \in \{ 1, \cdots, n \}$ (I'm considering the tensors $T$ and $S$ are $(0,2)$-tensors which has $n^2$ components). According this interpretation, we have in normal coordinates that

$\begin{align} (\circ) - H(\nabla_j h^i_k + \nabla_k h^i_j - \nabla^i h_{jk}) &=& - H \nabla_j h^i_k - H \nabla_k h^i_j + H \nabla^i h_{jk}\\ &=& \text{tr}_g A (\nabla_j h^i_k) + \text{tr}_g A (-\nabla_k h^i_j) + \text{tr}_g A \nabla^i h_{jk}\\ &=& \text{tr}_g A (\nabla_j g^{ir}h_{rk}) + \text{tr}_g A (-\nabla_k g^{ir}h_{rj}) + \text{tr}_g A \nabla^i h_{jk}\\ &=& g^{ir} \text{tr}_g A (\nabla_j h_{rk}) + g^{ir} \text{tr}_g A (-\nabla_k h_{rj}) + \text{tr}_g A \nabla^i h_{jk} = A \ast \nabla A, \end{align}$

where the last equality is valid according my interpretation because all the terms of the linear combination has a product among the contraction of the tensor $A$ and a component of the tensor $\nabla A$.

Furthermore, we have

$\begin{align} (\circ \circ) -h^i_k \nabla_j H -h^i_j \nabla_k H + h_{jk} \nabla^i H &=& -h^i_k \nabla_j (g^{rs}h_{rs}) -h^i_j \nabla_k (g^{rs}h_{rs}) + h_{jk} \nabla^i (g^{rs}h_{rs})\\ &=& -h^i_k \left( g^{rs} \nabla_j h_{rs} \right) -h^i_j \left( g^{rs} \nabla_k h_{rs} \right) + h_{jk} \left( g^{rs} \nabla^i h_{rs} \right)\\ &=& -h^i_k ( \text{tr}_g \nabla_j A ) -h^i_j ( \text{tr}_g \nabla_k A ) + h_{jk} ( \text{tr}_g \nabla^i A )\\ &=& -(g^{ir}h_{rk}) ( \text{tr}_g \nabla_j A ) -(g^{ir}h_{rj}) ( \text{tr}_g \nabla_k A ) + h_{jk} ( \text{tr}_g \nabla^i A )\\ &=& A \ast \nabla_j A + A \ast \nabla_k A + A \ast \nabla^i A, \end{align}$

where the last equality is valid according my interpretation because all the terms of the linear combination has a product among the contraction of the covariant derivative of the tensor $A$ and a component of the tensor $A$.

I'm stuck here, because I don't know how to show that $(\circ) + (\circ \circ) = \nabla H \ast A + H \ast \nabla A$. I think I misunderstood the definition because I cannot conclude what I want, so my question it's how to write the linear combination $T \ast S$ in order to understand how $\ast$-operation is defined.

$\textbf{EDIT:}$

$\begin{align} -h^i_k \nabla_j H -h^i_j \nabla_k H + h_{jk} \nabla^i H - H(\nabla_j h^i_k + \nabla_k h^i_j - \nabla^i h_{jk}) &=& - \nabla_j (h^i_k H) - \nabla_k (h^i_j H) - \nabla^i(h_{jk} H)\\ &=& - \nabla_j (g^{ir} h_{rk} \text{tr}_g (A)) - \nabla_k (g^{ir} h_{rj} \text{tr}_g (A)) - \nabla^i (h_{jk} \text{tr}_g (A))\\ &=& -g^{ir}(\nabla_j (h_{rk})\text{tr}_g(A) + h_{rk} \text{tr}_g(\nabla_j A)) -g^{ir}(\nabla_k (h_{rj}) \text{tr}_g(A) + h_{rj} \text{tr}_g (\nabla_k A))\\ &-& (\nabla^i (h_{jk}) \text{tr}_g(A) - h_{jk} \text{tr}_g(\nabla^i A))\\ &=& \left( -g^{ir} \nabla_j (h_{rk}) -g^{ir} \nabla_k (h_{rj}) - \nabla^i (h_{jk}) \right) \text{tr}_g(A)\\ &-& g^{ir} h_{rk} \text{tr}_g(\nabla_j A) - g^{ir} h_{rj} \text{tr}_g (\nabla_k A) - h_{jk} \text{tr}_g(\nabla^i A)\\ &=& \left( -g^{ir} \nabla_j (h_{rk}) -g^{ir} \nabla_k (h_{rj}) - \nabla^i (h_{jk}) \right) \text{tr}_g (A)\\ &-& g^{ir} h_{rk} \nabla_j (\text{tr}_g A) - g^{ir} h_{rj} \nabla_k(\text{tr}_g A) - h_{jk} \nabla^i(\text{tr}_g A)\\ &=& \text{tr}_g^{45} (\nabla A \otimes A) + \text{tr}_g^{45} (A \otimes \nabla A) = \nabla A \ast A. \end{align}$