You can also look what happens in the first part of the incline as a rotation:

[![Rotation.][1]][1]

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 $mg\sin\theta$.
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**Edit: Centripetal Force**

Firstly, $v^2=v_0^2-2gR(1-\cos\theta)$ can be easily verified by multiplying both sides with $\frac{m}{2}$ and reworking:

$$\frac{mv_0^2}{2}-\frac{mv^2}{2}=mgR(1-\cos\theta)$$

This is the energy conservation equation, e.g. for $\theta=\frac{\pi}{2}$ then:

$$\frac{mv_0^2}{2}-\frac{mv^2}{2}=mgR,$$

which is what we expect.

We know that $F_c=\frac{mv^2}{R}$, so that:

$$F_c=\frac{mv_0^2}{R}-2mg(1-\cos\theta)$$


  [1]: https://i.sstatic.net/6dYbr.png