Local frame formula for covariant differentiation along a curve

Apply the pullback prescription to $X={\mathrm{\partial }}_t$: its action on $v$ is $v^{\mathrm{\prime }}$ and the pulled-back one-form evaluated on ${\mathrm{\partial }}_t$ is $B(t)$. This gives the formula, and the frame is a basis, so zero covariant derivative is equivalent to $v^{\mathrm{\prime }}+Bv=0$.

Explicitly, for a second frame put $C(t)=A(\gamma (t))$, write $v=Cw$, and use $B^{\mathrm{\prime }}=C^{-1}BC+C^{-1}{C}^{\mathrm{\prime }}$ for the new connection matrix (here the prime on $B^{\mathrm{\prime }}$ denotes the new matrix). Then $v^{\mathrm{\prime }}+Bv=C^{\mathrm{\prime }}w+C{w}^{\mathrm{\prime }}+BCw=C(w^{\mathrm{\prime }}+B^{\mathrm{\prime }}w)$. Multiplication by the respective frames gives the same derivative. This computation allows singular curve velocity, including zero velocity, and uses no inverse of $\dot{\gamma }$. Rank zero means empty vectors; rank one is the scalar equation. Endpoint derivatives are one-sided, so the same product rule applies.