So here is the basic question I need you to answer in your head: is orbit a special case of free-fall? I think that will cut right to the essence of the question!
We grew up on this planet. We like to think about things happening on this planet. It's very easy to forget that space exists, that the Earth is round, that gravity eventually diminishes. If you do that, there is an acceleration g which varies depending on where you are, $g = 9.80\pm0.03 \text{ m/s}^2,$ and in the flat approximation free-fall from some origin is described by the equations
$$\begin{align}
x(t) &= u_0~t,\\
y(t) &= -\frac12 g t^2 + v_0 t
\end{align}$$
for some parameters $u_0$, $v_0$. Since it sounds kind of weird to say that something is falling when it is traveling upwards, we often stipulate that $v_0 = 0$ and the physical trajectory is $y(x) =- \frac{g}{2 u_0^2} x^2$.
A comparison to a circle with radius $r$, $$y=\sqrt{r^2 -x^2} = r\sqrt{1 -(x/r)^2}\approx r - \frac1{2r} x^2,$$
lets us say that at $t=0$ the radius of curvature of this trajectory is $r = u_0^2/g.$ And so our assumption that the Earth is flat is really an assumption about $r\ll R$ which is really an assumption about how fast things are going sideways, $u_0\ll \sqrt{g R}.$ Just to give you an intuition this speed is at Earth's surface 7.9 km/s or 28,000 km/hr or 18,000 mph or Mach 23. So in terms of your everyday experience with things moving less than half the speed of sound, this is a reasonable approximation. On the flip side I have heard people say (possibly inspired by this XKCD? Uncertain source) that “space isn’t far, space is fast,” referencing these same speeds.
But at your most extreme, a high enough sideways velocity causes orbit, you “fall” over a short time exactly as much as the curved surface of the Earth is falling away under you due to its own curvature, and thus you stay at a fixed distance from the surface. What does this look like in our flat-Earth approximation, where we pretend the surface does not curve away? As the surface actually accelerates downwards away from you, if you pretend that it is flat then there is an apparent (i.e. not actual) force drawing you upwards with magnitude $m v_x^2/R$. It scales proportional to $m$ so it appears to be an attenuation of $g$, it hits all masses in your jet airplane (or rocket or whatever) proportional to their mass just like gravity does. In the flat approximation, “orbit” becomes moving at Mach 23 so that this upwards drift acceleration cancels $g$ out completely. You might say that this is not “free-fall” because of this upwards drift force, maybe.
Generally physicists still call these orbits a special case of “free-fall,” but we are acknowledging that the Earth is round and thinking of it from an inertial point-of-view, looking down from outer space. So in this context all we see is your acceleration towards the surface which is due entirely to gravity. But if you’re looking at the same thing from the Earth’s surface it looks like the acceleration is zero, so something must be canceling out the direct effects of gravity.
Here is the last fact that you need to know about this situation. You probably thought that you were traveling at $v_x = 0$ right now as you “sit still” and read this. You are actually traveling at around Mach 1 in the direction that you instantaneously call East. The actual amount is more like Mach 1.35 times the cosine of your latitude, and it causes an upwards drift force that has the magnitude of $\cos^2\theta~0.034\text{ m/s}^2$ which has already been factored into that above $9.80 \text{ m/s}^2$ number for $g$. This is because every day you complete an orbit around the center of the Earth, traveling a distance of about $2\pi R \cos\theta$ per day. So this affects you, but only a little bit, attenuating $g$ by 0.3%. To first approximation, we can still speak about freefall parabolas on Earth’s flat surface, just attenuating $g$ appropriately. (Control question: this attenuation by 0.3% seems awfully small, no? But why—what do you think would happen if it were 80% or 100%?)
So what we are left with is that unfortunately both are right, in different contexts. If you are in the inertial context, staring from space at a rotating Earth, then yes, you will define free fall purely by the gravitational field which diminishes like the inverse square of the distance from Earth’s center. But if you are in the surface context, watching planes and trains and automobiles moving around nearby, then you are very likely to want to describe free-fall with an acceleration that (a) includes rotation and mass discrepancies in Earth and anything else that affects the local falling of a ball, and (b) either does not diminish at all with height, or for very high flying planes or balloons, maybe one that diminishes linearly with height. So your definition of free-fall acceleration needs to include these terms because your surface is accelerating but you want to conceptualize it as fixed.
