Let $G < Gl(n)$ be a Lie group, and let $g:\mathbb{R}^n\rightarrow G$ be a smooth curve, parametrized as $g(q(t))$ with $t\in\mathbb{R}$. In that case it holds that
$$ \tag 1 \frac{dg}{dt} g^{-1} \in \mathfrak{g} $$
Some computer vision papers, as well as some books (e.g. for reference, this) define the "left-jacobian" (which apparently differs from the concept of left derivatives on lie groups) of $G$ as
$$
\begin{equation}
J_l=\begin{pmatrix}
\left(\frac{\partial{g}}{\partial{q_1}}g^{-1}\right)^{\vee} & ... & \left(\frac{\partial{g}}{\partial{q_n}}g^{-1}\right)^{\vee} \\
\end{pmatrix}
\end{equation}
$$
where, supposing that we note $G_1,...,G_n$ the generators of $\mathfrak{g}$ , $\vee$ is defined as
$$ \vee : \mathfrak{g}\rightarrow \mathbb{R}^n $$ $$ \left(\sum_{i=1}^nx_iG_i\right)^{\vee}=(x_1,...,x_n)^T $$
and then they seem to deduce that
$$ \tag 2 \frac{dg}{dt}g^{-1}=\sum_{i=1}^nw_iG_i $$
with $w_i$ the $i$-th component of $W=J_l\frac{dq}{dt}$.
I fail to see why equation (2) is correct (because of $W$). Assuming that I can start from (1) by using the chain rule, i.e. writing $\frac{dg}{dq}\frac{dq}{dt}g^{-1}$, I am unsure about the way $\frac{dg}{dt}$ and $g^{-1}$ interact. I assume they do not necessarily commute, which confuses me even more, considering that $\frac{dg}{dt}$ ends up at the right-side of the expression of $W$.