I'm not sure what you intend by the terminology "locally flat injective loop". Do you intend
this in the sense of topology of manifolds, that it's a continuous injective map where
each point in the image has a neighborhood making the curve homeomorphic to a line in
If this is the correct interpretation, then it isn't true even for curves in the plane.
You can start with a circle, and make a countable sequence of embellishments in various
intervals by putting in an inward spiral and matching outward spiral of total
angle $4\pi$ each.
Parametrize it so that the inward spiral for the $i$th embellishment
takes a very short time $< \epsilon_i$, then it lingers in the
middle for some time say $> 10 \epsilon_i$, and then it spirals outward in a way that steps
of length $< 10 \epsilon_i$ traverse angles $< \pi/2$.
Then for any series of geodesic steps with parameter length between $\epsilon_i$ and $10 \epsilon_i$ are self-intersecting in this embellishment, because the turning number of the inward journey
does not match the turning number of the outward journey.
With a sequence of embellishments of this form where the $\epsilon_i$ form a geometric
sequence, you can make polygonal approximations
with steps of sufficiently small $\epsilon$ all have self-intersection. In the plane,
of course, if you choose $\epsilon$ large enough that there are only 3 points, you'll get
a triangle, and it will be simple. If you want to prevent even large polygons from
being simple, I'm confident you can do it with a different Riemannian metric on the plane
(and therefore for some Riemannian metric on any other surface and on any manifold of any dimenson).
As noted in comments, every injective image of a circle in the plane is locally flat. Here is a plot of a suitable sample embellishment. The blue curve is the Jordan curve, parametrized in a way that the inward spiral is traversed nearly 20 times as fast
as the outward spiral. For some range of stepsizes, the inward journey (starting from the left) is forced to intersect the outward journey, because the total angle of turning in the inward journey
is 0 (since there is just one intermediate step), while in the outward journey it turns counterclockwise two full revolutions.