I am no physicist, but I have thought about connections associated to submanifolds. I don't know how to describe the specific construction in your question, but it reminds me of the Bott connection associated to the foliation. When $M$ is pseudo-Riemannian and $\Sigma$ is nondegenerate, $\Sigma$ is locally the leaf of a foliation via the Riemannian structure. So, I will just deal with foliations $\mathcal{F}$ where $\Sigma$ is a leaf.
Connections associated to a foliation:
There is the Bott connection on the normal bundle of $\mathcal{F}$. The normal bundle $N_{M/\mathcal{F}}$ is defined in the same way as the normal bundle of a submanifold, but is on $M$:
$$N_{M/\mathcal{F}} = T_M/T_\mathcal{F}.$$
On the leaves of the foliation, the normal bundle has a canonical connection given by the Lie bracket:
$$\nabla^{Bott}: N_{M/\mathcal{F}} \otimes \Gamma(T_\mathcal{F})\to N_{M/\mathcal{F}},\\
\nabla^{Bott}_X Y = [X,Y] = \mathcal{L}_X(Y).$$
It is a nice exercise to check that this gives a well-defined connection.
This restricts to a connection on $N_{M/\mathcal{F}}|_\Sigma$. To compute $\nabla_X Y$ for $X \in \Gamma(T_\Sigma)$, first extend $X$ to $\tilde X$ tangent to the foliation in a neighborhood of $\Sigma$, extend $Y$, and then compute with the Bott connection. There is also a dual connection on the conormal bundle $N^\vee_{M/\mathcal{F}} = (T_\mathcal{F}^\perp \subseteq \Omega^1_M)$ over the foliation, again given by the Lie derivative.
I suspect that your object $\mathcal{L}_Nq$ is defined similarly, but using the orthogonal complement under the Riemannian structure instead of the normal or conormal bundles. It should be well-defined in the same way the Bott connection is well-defined.
I think your work shows that the directional derivative induced by the curve $\phi^{-1}\circ(\phi\circ \sigma_1+\phi\circ\sigma_2)$ does not depend of your choice of $\phi$, nor representatives of the equivalence classes of $\sigma_1$ and $\sigma_2$ (with $\sigma_1\sim\sigma_2$ iff there is a chart $(\phi,U)$ in p with $(\phi\circ\sigma_1)'(0)=(\phi\circ\sigma_2)'(0)$ $(\ast)$).
We know that tangent vectors and directional derivatives are the same through the linear isomorphism
$$v=[\sigma]\mapsto(f\mapsto (f\circ\sigma)'(0)),$$
but this latter already suppose we know that there is a vector space structure on the tangent vectors defined as equivalence classes of curves. So what we actually have to prove is that
$$\phi^{-1}\circ(\phi\circ \sigma_1+\phi\circ\sigma_2)\sim\psi^{-1}\circ(\psi\circ \tilde\sigma_1+\psi\circ\tilde\sigma_2)$$
for another $\psi:M\to\mathbb{R}^m$ s.t. $\psi(p)=0$, and $\sigma_1\sim\tilde\sigma_1,\sigma_2\sim\tilde\sigma_2$. Happily, the proof looks like a lot the one you already produced:
Choose $(\phi,U)$ as the map we will use to check $(\ast)$. The derivative of the first curve is
$$(\phi\circ\phi^{-1}\circ(\phi\circ\sigma_1+\phi\circ\sigma_2))'(0)=(\phi\circ\sigma_1)'(0)+(\phi\circ\sigma_2)'(0).$$
The linearity of the derivative is indeed checked here: for functions $F,G:\mathbb{R}\to\mathbb{R}^m$, we can write $F(t)=(F_1(t),\dots,F_m(t)),G(t)=(G_1(t),\dots,G_m(t))$ and since $(F+G)(t)$ is no more than $(F_1(t)+G_1(t),\dots,F_m(t)+G_m(t))$, the linearity is the one for functions from $\mathbb{R}$ to $\mathbb{R}$.
The derivative of the second curve is
\begin{align*}
&(\phi\circ\psi^{-1}\circ(\psi\circ \tilde\sigma_1+\psi\circ\tilde\sigma_2))'(0)\\
=\,&D(\phi\circ\psi^{-1})_0\left((\psi\circ\tilde\sigma_1)'(0)+(\psi\circ\tilde\sigma_2)'(0)\right)\\
=\,&D(\phi\circ\psi^{-1})_0\left((\psi\circ\tilde\sigma_1)'(0)\right)+D(\phi\circ\psi^{-1})_0\left((\psi\circ\tilde\sigma_2)'(0)\right)\\
=\,&(\phi\circ\tilde\sigma_1)'(0)+(\phi\circ\tilde\sigma_2)'(0)\\
=\,&(\phi\circ\sigma_1)'(0)+(\phi\circ\sigma_2)'(0),
\end{align*}
this last equality coming from $\sigma_i\sim\tilde\sigma_i$, $i=1,2$.
Best Answer
I need to assume that $g(\gamma(0)) = e$. Compute $$\frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} (\rho\circ g\circ \gamma)(t)v = {\rm d}\rho_{g(\gamma(0))} ((g\circ \gamma)'(0))v = \rho_\ast((g\circ \gamma)'(0))v = \rho_\ast(-\mathcal{A}(X))v = -\rho_\ast(\mathcal{A}(X))v.$$
It's very simple if you understand all the abuses of notation happening here.
That's all there is to it.