You can also look what happens in the first part of the incline as a rotation:
The ball rotates around the point $O$, with radius $R$.
The centripetal force $F_c$ is provided by the Normal force ($mg\cos\theta$), while $F_t=mg\sin\theta$ provides a decelerating torque, so that:
$$-Rmg\sin\theta=I\frac{d\omega}{dt}$$
$$-Rmg\sin\theta=mR^2\omega \frac{d\omega}{d \theta}$$
$$-R\omega d\omega=g\sin\theta d\theta$$
Note that the ball enters the incline at $v_0$, so $v_0=\omega_0 R$.
Integrating between $0$ and $\theta$ we get:
$$\omega^2-\omega_0^2=\frac{2g}{R}(\cos\theta -1)$$
Re-worked with $\omega=\frac{v}{R}$, we get:
$$v^2=v_0^2-2gR(1-\cos\theta)$$
So $v$ is reduced and the vector now also has a vertical component:
$$v_y=v\sin\theta$$
Once the ball leaves the arc the only net force acting on it is $mg\sin\theta$.