I want to show $(2.3.11)$, but in my calculation, there is no minus.
What I try: from the $(2)$, I have $$ h(X,\text{Ric}(W))=h(X, \text{Ric}(W, e_i)e_i)= \text{tr}\space h( X, \text{Ric}(W, \cdot)\cdot)=\text{tr}\space h(\text{Ric}(W, \cdot)\cdot,X) $$ The last equality is from $h=\partial_t g$ where $g$ is Riemannian metric (namely $h$ is symmetric).
Therefore, I think there should be no minus in the red line. But I am not sure, so, ask here.
It will important during the proof to be able to juggle different formu-lations of the lower order terms in the definition of $\Delta _L$. First, we have: \begin{align} \tag{2.3.11}h\left(X,\text{Ric}\left(W\right)\right)&=\langle h(X,\cdot),\text{Ric}(W,\cdot)\rangle \\ &=\text{tr}\space h(X,\cdot)\space\otimes\text{Ric}(W,\cdot) \color{red}{\underline{\color{black}{=\text{-tr}\space h(R(W,\cdot)\cdot,X),}}} \end{align} which one can easily check with respect to an orthonormal frame {$e_i$}; for example : \begin{align} \tag{2}\text{tr}\space h(X,\cdot)\space\otimes \text{Ric}(W,\cdot)&=\sum _i h(X,e_i)\text{Ric}(W,e_i)\\ &=h(X,\text{Ric}(W,e_i)e_i)\\ &=h(X,\text{Ric}(W)). \end{align}
I have made your mistake too. The reason is simple. In Topping's Book, $\mathrm{Ric}(X,Y)$ has been defined as $\mathsf{tr} Rm(X,.,Y,.)$ and this is equal to $-\mathsf{tr} Rm(X,.,.,Y)=-\mathsf{tr} \langle R(X,.).,Y\rangle$ by skew-symmetric properties of Riemann curvature tensor; and I think the rest is clear? (Be careful that in 2.3.11 it is $R$ and not $\mathrm{Ric}$.)