Let $M$ be a smooth manifold, $V$ a finite-dimensional real vector space, $\omega\in\Omega^1(M)\otimes\mathrm{End}(V)$ and $F\in\Omega^2(M)\otimes\mathrm{End}(V)$. Furthermore we are given a chart $x$ on $M$ that maps $x_0\in M$ to $0$ and we consider the vector field $R:=x^i\partial_i$. Lastly, we suppose that \begin{equation}\tag{1.12} L_R\omega=\iota_RF. \end{equation} From the proof of Proposition $1.18$:
Expanding the Taylor's series of both sides of $(1.12)$, we obtain $$\sum_\alpha(|\alpha|+1)(\partial^\alpha\omega_l)(0)\frac{x^\alpha}{\alpha!}=\sum_{\alpha,k}\partial^\alpha F(\partial_k,\partial_l)_{0}x^k\frac{x^\alpha}{\alpha!}$$
I think that on the RHS $\partial^\alpha(F_{kl})(0)x^k$ must be replaced by $\partial^\alpha(F_{kl}x^k)(0).$ Here is my reasoning:
On the one hand $$L_R\omega=L_R(\omega_i\wedge\mathrm{d}x^i)=(L_R\omega_i)\wedge\mathrm{d}x^i+\omega_i\wedge\underbrace{(L_R\mathrm{d}x^i)}_{=\mathrm d(R^i)=\mathrm{d}x^i}=(\omega_i+R\omega_i)\wedge\mathrm{d}x^i$$ and on the other hand $\iota_RF=F(R,\partial_i)\mathrm{d}x^i$ such that $(1.12)$ is equivalent to the following proposition: $$\forall i:\omega_i+R\omega_i=F(R,\partial_i)$$ In addition $\forall\phi:\partial^\alpha(R\phi)(0)=|\alpha|(\partial^\alpha\phi)(0)$ and hence we obtain the following result: $$\forall\alpha:(|\alpha|+1)\partial^\alpha\omega_l(0)=\partial^\alpha(\omega_i+R\omega_i)(0)=\partial^\alpha F(R,\partial_i)(0)=\partial^\alpha(F_{ki}x^k)(0).$$