Suppose we have a principal bundle $p: P \to M$ with principal connection form $\omega: TP \to \mathfrak{g}$. 


Given any arbitrary curve $\gamma: I \to P$ in the total space of the principal bundle, where $I=[0,1]$, we can use the nonabelian fundamental theorem of calculus [[R. W. Sharpe](https://books.google.co.nz/books?id=Ytqs4xU5QKAC&pg=PA96&lpg=PA96&dq=non+abelian+fundamental+theorem+of+calculus&source=bl&ots=xritYLD8wq&sig=34nrRs_UM12MaYwrRJatX44xy2M&hl=en&sa=X&ved=0ahUKEwiwgaSdwe3RAhWHE5QKHZyoDzcQ6AEIHjAB#v=onepage&q=non%20abelian%20fundamental%20theorem%20of%20calculus&f=false) Thm 7.14] to integrate $\gamma^\*\omega: TI \to \mathfrak{g}$ to a map $g: I \to G$ such that $g(0) = e$ and $g^\* \omega\_G = \gamma^\* \omega$, where $\omega\_G: TG \to \mathfrak{g}$ is the usual Maurer-Cartan form on $G$.


Now we can define a sort of transport of the starting point $\gamma(0)$ along this curve by
\begin{equation}\gamma(0) \mapsto \gamma(t).g(t)^{-1}.\end{equation}
**Note:** I might need to replace $g(t)^{-1}$ by $g(t)$ in the formula above.


On the other hand, we can use the standard definition of parallel transport: take the projection $p \circ \gamma: I \to M$, and lift it again to a horizontal curve $\widetilde{\gamma}: I \to P$ with the same starting point $\widetilde{\gamma}(0)=\gamma(0)$. Then we can parallel transport the starting point $\widetilde{\gamma}(0)=\gamma(0)$ by traveling along the horizontal curve:
\begin{equation}\mathrm{Pt}\_{p \circ \gamma}(t)\gamma(0)= \widetilde{\gamma}(t). \end{equation}


Do these two notions of transport agree? In other words, do we have an equality $\widetilde{\gamma}(t) = \gamma(t).g(t)^{-1}?$ 


For anyone not familiar with the nonabelian FTC, I'll give the global version here:


**Theorem (nonabelian fundamental theorem of calculus):** Suppose we are given a connected manifold $M$ equipped with a $\mathfrak{g}$-valued $1$-form $\omega: TM \to \mathfrak{g}$. Then there exists a smooth map $f: M \to G$ such that $\omega =f^\* \omega\_G$ if and only if $\omega$ satisfies the Maurer-Cartan equation $d\omega + \frac{1}{2}[w \wedge \omega]=0$ and the monodromy representation $\pi\_1(M, x) \to G$ is trivial. Moreover, if these conditions are satisfied then the map $f$, called the *integral* of $\omega$, is unique up to left translation by an element of $G$.


**Corollary:** On the interval $I=[0,1]$ any 1-form $\omega: TI \to \mathfrak{g}$ has an integral $f: I \to G$ unique up to choice of $f(0) \in G$.



Parallel transport in a principal bundle via the nonabelian fundamental theorem of calculus
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 $A \in \mathfrak{g}$.


