Great question. If I could somewhat reframe your question, I believe you confused by the apparent contradiction between these two statement about the equivalent principle:
- Any manifold in GR locally looks like Minkowski space.
- Even (very small) local frames can demonstrate gravitational effects (e.g. you can feel yourself "accelerating upward" if you are standing on the surface of the Earth).
Your objection is that the statement 1 seems to imply that there can't be any observable gravitational effects over very small regions of spacetime, while statement 2 seems to imply that there can be.
The resolution to this apparent contradiction is that statements 1 and 2 and using different quantitative notions of the word "local", and statement 1 is restricting the word "local" to smaller regions than statement 2 is.
More precisely: statement 1 can be more precisely rephrased as:
For any point $p$ on any pseudo-Riemannian manifold (i.e. spacetime), there exists a local coordinate system around $p$ in which the Taylor expansion of metric tensor agrees with the Minkowski metric $\eta$ to first order about $p$.
In other words, $g(p) = \eta$ and $\partial_\mu g(p) \equiv 0$ in these particular coordinates (which are known as Riemann normal coordinates). So if you define "local" to mean "so small that only first-order variations are non-negligeable," which is the implicit assumption in statement 1, then indeed no gravitational effects can be detected locally.
But it turns out that the effects of curvature (or the acceleration of the "owner" of a local coordinate system) necessarily come in at second order in the metric. More precisely:
A manifold has intrinsic curvature at a point $p$ iff the second-order Taylor expansion of the metric tensor about $p$ deviates from the Minkowski metric.
Or even more precisely:
At any point $p$ on any pseudo-Riemannian manifold, the second-order partial derivatives of the metric tensor $\partial_\mu \partial_\nu g(p)$ are either identically zero in every coordinate system, or have some nonzero components in every coordinate system. Therefore, the proposition $\partial_\mu \partial_\nu g(p) \equiv 0$ is coordinate independent. The Riemann curvature tensor vanishes at $p$ iff $\partial_\mu \partial_\nu g(p) \equiv 0$ in some coordinate system (and therefore in all of them).
Therefore, you can always make the Taylor expansion of the metric about a point agree with the Minkowski metric to first order (by using Riemann normal coordinates), but you can't make it agree to second order if the manifold is curved at point $p$. Since gravitational effects are a physical manifestation of the curvature of the spacetime manifold, you can detect them if your local frame is big enough to capture second-order deviations about the point $p$. This slightly weaker sense of "local" is the sense being used in statement 2. If your region of spacetime is only "first-order big" in the time direction, then you won't have time to measure any relative acceleration of a nearby test particle.
(By the way, you can't actually forces that induce an acceleration, but only forces that induce a tidal acceleration - defined broadly as any spatial variation in the acceleration field. The only reason that you can feel the Earth accelerating you upward is because your body is big enough that the second-order terms in the metric (which are proportional to the acceleration constant $g$) are non-negligeable. You many not be used to thinking of the electrostatic repulsion acceleration of the Earth pushing up on you as a "tidal" acceleration, but it is: the only reason you can feel it is that it's applied at the soles of your feel but not elsewhere on your body, which induces internal compression forces within your body that you feel. If it were somehow distributed in such a way as to induce a uniform acceleration over your entire body, then it would work just like gravity and you wouldn't be able to feel it.)
