When a horse pulls a cart, by Newton's III law the cart also pulls the horse with equal and opposite force, then how does the motion of horse, cart and system takes place?

Explanation

Here the following forces acts along the horizontal,
(1) The force on the cart due to the pull of the horse `F_(CH)`
(2) The force on the horse due to the cart `F_(HC)`
(3) The horizontal force on the cart due to ground f.
(4) The reaction from ground due to the push of horse against the ground `F_(HG)`
The vertical forces on the cart and horse are their weights vertically down while reactions due to contact from ground upward. This all is shown in

So, for the motion of the cart,
`F_(CH) gt f` .....(i)
i.e, the force with which the horse pulls the cart `(F_(CH))` is responsible for the motion of the cart and the cart will move if this force exceeds the static friction (f) between the cart and the ground.
And for the motion of horse
`F_(HG) gt F_(HC)` ....(ii)
i.e, the force which the ground exerts on horse (i.e., `F_(HG)`) is responsible for the motion of the horse and the horse will move if this force exceeds the force with which the cart pulls the horse `(F_(HC))`
And so for the motion of the system [adding Eqns. (i) and (ii)]
`F_(HG)+F_(CH) gt F_(HC)+f`
or `F_(HG) gt f [as F_(CH) = F_(HC)]` ....(iii)
i.e, the force which the ground exert on the horse (due to its push against the ground) is responsible for the motion of the system. Free body diagrams for cart, horse and system are shown in

Note: Same argument is applicable in tug of war. The team that pushes harder against the ground wins. The tension of the rope pulls a team backward. it is the earth's reaction to the push against the ground which supplies the necessary force for pulling.