Does the presence of high level trainers have an effect on low level ones?

Consider a local Lorentz frame, $g_{ij} = \mathrm{diag}(1,-1,-1,-1)$. An observer (really a congruence of observers) at rest with respect to this frame has velocity vector $u^i = \delta^i_0$. It experiences no force (geodesic deviation) if it obeys the geodesic equation, which in this case becomes simply $\gamma^i{}_{00} \equiv 0$. From the compatibility of the connection with the metric we know that $\gamma^i{}_{00} \equiv 0$ if and only if $\gamma^0{}_{i0} \equiv 0$, and from the first Cartan equation $$ d\omega^i = \omega^j \wedge \gamma^i{}_j = \gamma^i{}_{jk}\omega^j\wedge\omega^k, $$ we know that this is certainly the case if $d\omega^0 \equiv 0$. Here $\gamma^i{}_j$ are the connection forms and $\gamma^i{}_{jk}$ are the components (Ricci rotation coefficients). Given a set of coordinates it is natural to consider an observer to be static if the velocity vector is given by $$ u^\mu = \frac{1}{\sqrt{g_{00}}}\delta^\mu{}_0 $$ (here I let $\mu,\nu,\ldots$ signify coordinate indices). Considering then the static case ($g_{0\mu} \equiv 0$ for all $\mu = 1,2,3$) we find that upon setting $e^\mu_0 = u^\mu$ we have $d\omega^0 \equiv 0$ whenever $g_{00}$ is constant. Such is the reasoning behind the statement that gravitational attraction arises from non-constant $g_{00}$. As you can see it is most definitely a simplification.

As is apparent from the above exposure, any observer not at rest with respect to our local Lorentz frame may experience a force (though the nature of its correspondence to high speeds depends on the exact form of the metric and/or connection forms).

As to your questions regarding the effects on light, it is important to recall that light follows null geodesics. Therefore they will always be affected by the nature of the rotation coefficients $\gamma^i{}_{00}$ but also by at least some other coefficients. It would require a speed greater than that of light (spacelike observer) to escape to effects of $\gamma^i{}_{00}$, but this is clearly unphysical.

Although, as John Rennie links to in his answer, it is meaningless to talk about curvature in one direction, in light of the above considerations we might ponder the case where $\gamma^i{}_{00} = \gamma^0{}_{i0}$ are the only non-zero rotation coefficients. This corresponds concretely to the simplest case of the greater the speed with respect to our frame, the smaller the "curvature effects" on the motion (though as noted above, one would require a speed greater than that of light to escape them altogether). Then $d\omega^i \equiv 0$ for all $i = 1,2,3$. By the second Cartan equation $$ d\gamma^i{}_j = \gamma^k{}_j\wedge\gamma^i{}_k + \frac{1}{2}R^i{}_{jk\ell}\omega^k\wedge\omega^\ell, $$ we immediately find $$ d\gamma^0{}_i = -\gamma^0{}_{i0|j} \omega^0 \wedge \omega^j = R^0{}_{i0j} \omega^0 \wedge \omega^j $$ to give the only (potentially) non-zero curvature components, up to symmetries. Note that we take $i,j \neq 0$, whence in particular it follows that the Ricci tensor is zero if and only if the Riemann tensor is. Therefore we can at least conclude that such solutions cannot be vacuum, and therefore cannot describe the exterior of any object.

EDIT: In fact, I was a bit lazy in concluding the above. Making the contractions, and ignoring any cosmological constant, we find that the Einstein field equations yield $T_{0i} = 0$ for all $i$, whence any (non-flat) solution must violate the dominant energy condition. Thus we can further conclude that such a solution is unphysical, since there are timelike observers that observe energy to flow faster than the speed of light, i.e. timelike vectors $v^i$ such that $T_i{}^jv^i$ is spacelike (namely all observers not at rest with respect to our frame).