Yes, they coincide (at least if I understand you correctly).

Let $\check{\gamma}$ be a path in $M$. If we fix a lift $\gamma$ in P, then the horizontal lift $\gamma_\omega$ of $\check{\gamma}$ is necessarily of the form $\gamma_\omega(t) = \gamma(t) \cdot g(t)$ for some curve $g: [0,1] \to G$.
Differentiating this relation with respect to $ t $ yields 
$$
\omega(\dot \gamma_\omega(t)) 	= \omega(\dot \gamma (t) . g(t)) + \omega(\gamma(t) . \dot g(t)) \\
=  \omega(\dot \gamma (t) . g(t)) + \omega(\gamma(t) \cdot g(t) . g(t)^{-1} \dot g(t)) \\
= Ad^{-1}_{g(t)} \omega(\dot \gamma (t)) + g(t)^{-1} \dot g(t).$$
Using the fact that that $\gamma_\omega$ is horizontal, we consequently get $\dot g(t) g^{-1}(t) = - \omega(\dot \gamma (t))$.

A few words about the notation used in the above calculation:

- The lower dots are just the partial derivatives of the the action. In particular $p . A = \frac{d}{d \epsilon} (p \cdot exp(\epsilon A))$ for $A$ a Lie algebra element and $p \in P$.
- Equivariance of the connection implies $\omega(X . g) = Ad^{-1}_g \omega(X)$ for every $X \in T_p P$.
- The connection is the identity on vertical vector, which amounts to $\omega(p. A) = A$ for every $p \in P$ and $A \in \mathfrak{g}$.