How to calculate the weight my DC motor can move?
I think you don't fully understand the specs of the motor.


  Stall Torque = 20 Kg/cm


The unit is Kg*cm. And when a wheel of 4" diameter, i.e. 2"=5cm radius is directly mounted to the shaft, the "force" is 20Kg*cm / 5cm = 4kg.
This means: If you put a string around the wheel and use it as winch, it could lift a weight of almost 4kg (8.8lbs) before it stalls.


  Motor Rated RPM = 133


If there is no weight attached, the wheel would spin with 133RPM. The circumference of the wheel is 2*pi*5cm=31.4cm (~1foot), so the string would move up with 133ft/min=40.5m/min=0.8m/s=2.4km/h. This would also be the absolute top speed of your cart. In reality, the speed will decrease when a load is applied.

Let's go a step further and calculate acceleration. Let's assume a constant, max torque and a mass of 100kg.

$$F=m_{force}\cdot g=m_{cart}a$$
$$\Rightarrow\quad a=\frac{4kg}{100kg}\cdot 9.81m/s^2=0.4m/s^2$$

With this acceleration, the car would reach its top speed within 2 seconds. Not bad. 



Now, as said in the comments, there's lots of friction. Friction in the bearings, friction between wheels and ground (Asphalt? Grass? Sand?) And when the cart makes a turn and only one inner wheel is driven by the motor, this increases the necessary force drastically. The motor has to overcome all of this.

I'd first measure how much "force" is needed to move the cart. Let someone sit in the cart, and push it with a kitchen scale. Is 4kg always enough to move? (Acually, you should need less than that)

Finally, I'd guess it is not enough to reliably move the cart. But since you somehow have to connect the motor to the wheel, a gear with a ratio of 2:1 would half the top speed, but double the force.

And one very important point: The gearbox already attached to your motor is made to withstand the forces during normal operation plus some margin. So it is made to handle may be 30-40kg*cm. But if someone jumps into the cart, this could simply crush the gearbox!