If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the Earth's radius, $g$ can be taken to be constant, equal to $9.8~\textrm{m/s}^2$. Free fall is thus a case of motion with uniform acceleration.
What is meant by: If the height through which the object falls is small compared to the Earth's radius?
So you probably learned two equations for gravitational force, the general one being \begin{equation} F_g = G \frac{m_1 m_2}{r^2}. \end{equation}
Then there is the one that we use only on earth's surface \begin{equation} F_g = m g. \end{equation}
The one for earth's surface is actually the same as the generic one. The thing we did was setting $G\frac{m_1}{r^2} \sim g$, with $m_1$ being the Earth's mass. We can do this since $G$ is a constant, the earth's mass $m_1$ is a constant, and the radius to the core $r$ is roughly a constant (measured from Earth's surface).
So if you keep in mind what we just called $g$, you can see that if the radius to the core $r$ is not constant (you increase your distance to Earth's core by going to space for example), then the approximation for $g$ breaks down.
This is why the second formula only works on earth's surface and gives you false results when you move away from it.
With Newton law and gravitation force you obtain
$$m\,g=F=G\,\frac{m\,M}{r^2}\tag 1$$
at the earth surface is $r=r_E$ where $r_E$ is the earth radius,$M$ is the earth mass and $G$ the gravitation constant for $g$ you obtain
$$g=G\,\frac{M}{r_E^2}\tag 2$$
but if your position is $r\mapsto r_E+h$ where h is your height from the earth surface , you obtain
$$g=G\,\frac{M}{(r_E+h)^2}=G\,\frac{M}{r_E^2\,(1+h/r_E)^2}$$
because $h/r_E\ll 1 $ we can neglected this part and obtain the result of equation (2)
lets put some numeber.
earth radius $r_E\approx 6371 ~$[km] and $~h=10~$[km] so you obtain $h/r_E~=0.0015\ll 1$
With distance, g lessens according to the inverse square law. this can usually be assumed to be negligible near the Earth's surface where, compared to Earth's radius, the distance it falls is small. Farther from Earth we must take into account that g will be less than 9.8 mps^2 by the inverse square law. Hope this intuitive explanation helps.
Note the mention that "air resistance is neglected". This is a hint that statement is valid only at or near the Earth's surface.
You are right, the radius of the Earth does not matter. What does matter is that the height through which the object falls is small compared to the object's distance from the Earth's center. But if we consider that the object is in the Earth's atmosphere, then it's distance from the Earth's center is very close to the Earth's radius. The statement was poorly worded.
Perhaps a more general statement would have been:
...If the height through which the object falls is small compared to the distance from the Earth's center, g can be taken to be constant...
