Locally flat coordinate and Locally inertial frame

In the most general case described by general relativity, it is not possible to find a $neighbourhood$ covered by coordinates $x^\mu$ such that $g^{\mu\nu} = \eta^{\mu\nu}$ in all U. If it were so, you have a zero Riemann tensor, hence the space-time would be flat in all U. You may have space-times with such flat pieces (I think there is no problem in gluing this piece with non-flat pieces, but I may be wrong), but is not the most general case and not what is meant when we say that space-time is locally flat.

What we mean is that the tangent space, in any point is the Minkowski space-time.

This mean that, for any point p, you can find a basis for the tangent space at p (and associated "exponential" coordinates) so that the metric is diag(-,+,+,+) in these coordinates at this point p and the connession coefficients vanish at this point (not in a neighbourhood!)

You can think of these coordinates as those of a inertial observer. Note that there exists several possible coordinates, which are related by a Lorentz transform at the tangent space, and are associated to different observers.

In what sense you can think of these coordinates as those of a inertial observer? In the sense that as long as you are covering a sufficiently small neighbourhood of p, whose dimension will be "smaller the larger the Riemann tensor is at p", you may describe everything happening here as if you were in special relativity. One above all, the geodesics are of the form $d/dt^2 x(\tau) = 0$ and do not accelerate with respect to each other. Of course, actually they do, but these effects are small if you consider small neighbourhood of p and small Riemann at p.

Analogously, Earth is flat at a point in the sense that you can "confuse" the flat tangent space with the actual neighbourhood because the differences are difficult to detect if you zoom enough.

In light of our clarifying discussions, I believe the answer is yes.

I found a nice section on Fermi Normal coordinates here (Section 9):

http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html

This seems to be what you mean by "Locally inertial coordinates" - the tetrad is orthonormal with one direction along the curve, and the others along spacelike curves orthogonal to the curve.

Since you can define Fermi normal coordinates anywhere on a timelike geodesic, define them on the intersection of two geodesics. These define a flat metric, so there's no reason why you couldn't pick that metric to be the tetrad for the other observer at the same point.

If the observer is not in free-fall, the metric-tensor $g_{\mu,\nu}(s)$ at the observer's position, expressed in local coordinates around the observer, will not be $\eta_{\mu,\nu}$. Your first assumption about the path $(\gamma)$ is wrong.

I guess what you are aiming at is the notion of the space of coordinates around a point, which is indeed a flat space (since it is (pseudo-)euclidean). This space however serves only to introduce coordinates in an open subset of your manifold by a mapping that is homeomorphic to an open subset of that (pseudo)-euclidean space. This means that the open subset of your manifold is quite the same (up to deformations, that is curvature!!!) as the (pseudo-)euclidean subset.