Dear Steve Huntsman, I find that the notion of fundamental vector field is well defined not only for $G$-principal bundle but even for any $G$-manifold, i.e. a manifold with an action by a Lie Group.
About your feeling of unsatisfaction with the definition, and your notional equation, I would say that effectively the fundamental ve
Let $\Psi:M\times G\to M$ be a right action of a Lie group $G$ on a manifold $M$. Let $\frak{g}$ be the Lie algebra of $G$, viewed as formed by the left invariant vectorfields on $G$. Then there exists a unique map $\zeta:X\in\frak{g}\to \zeta_X\in\frak{X}(M)$ such that $(T\psi)\circ(0_M+X)=\zeta_X\circ\Psi$, $\zeta_X$ and $0_M+0_M$ are $\Psi$-related, for any $X\in\frak{g}$.(Above $0_M$ denoted the zero vectorfield on $M$.) Obviously the following property is satisfied: $\zeta_{aX+bY}=a\zeta_X+b\zeta_Y,\zeta_{[X,Y]}=[\zeta_X,\zeta_Y]$, for any $a,b\in\mathbb{R}$, and $X,Y\in\frak{g}$, i.e. $\zeta:\frak{g}\to\frak{X}(M)$ is a Lie algebra homomorphism. For any $X\in\frak{g}$, the vector field $\zeta_X$ on $M$ is called the fundamental vectorfield corresponding to $X$ w.r.t. the right action $\Psi$. Remark: In order to justify the definition of $\zeta$ we have only to remark that, for any $X\in\frak{g}$, the map $T\Psi\circ(0_M+X)$ is constant on the fibers of $\Psi$, thanks to the left invariance of $X$.