The two cases are different. Take a car rolling along a road that encompasses
a great circle of the Earth (e.g. the equator, if such a road could be built). We will assume that the car is not accelerating $along$ the road (speedometer gives a constant reading). We will also neglect the rotation of the Earth in this example.
From the viewpoint of an outside inertial observer, the car will necessarily follow a curved path to match the curvature of the Earth. The equation of motion in the radial direction is:
$$\Sigma F=ma_r$$
$$N-W=-m\frac{v^2}{r}$$
or expressing weight using Newton's law of gravitation:
$$\frac{GM}{r^2}-\frac N m=a_r$$
Here, the normal force $N$ and the weight $W$ do not sum to zero, and there is a non-zero acceleration $a_r$ that depends on the speed $v$ of the car. Note that if the car's speed is reduced to $v=0$, the acceleration is also zero (and the accelerometer placed in the car would record 0 g).
Now consider the case of a satellite in circular orbit. In the inertial frame there is only $one$ force acting: the weight.
$$\Sigma F=ma_r$$
$$-W=-m\frac{v^2}{r}$$
The acceleration $a_r$ in this case is then, per Newton's law of gravitation:
$$m \frac{GM}{r^2}=ma_r$$
$$\frac{GM}{r^2}=a_r$$
or, for a radius negligibly different than the Earth's surface, $a_r= 1$ g (downward).
This is what an accelerometer in the satellite will read.
These are the accelerometer readings you will get if you tare the device so it reads 0 g while sitting unmoving on the ground, and reads 1 g (downward) when in free fall. This convention is in line with Newtonian mechanics. You can also tare the device to read 0 g in free fall and 1 g (upward) while stationary on the ground, which is equivalent but more in line with General Relativity. But Relativity is not necessary to understand or predict the motions outlined here.
I encourage you to draw the free body diagrams and try to derive these equations as shown.