Your sketch is misleading. The disk is **negligibly small as compared with the sphere**. Close to the surface the disk and sphere are flat and the electric fields of both are uniform.

The electric field at the surface of a uniformly charged spherical shell is the same as from a point charge : $E=\frac{Q}{4\pi \epsilon_0 R^2}=\frac{\sigma}{\epsilon_0}$ where $\sigma$ is the surface charge density. Moreover, it has this value immediately outside the shell, and is $0$ inside the shell (see **Newton's Shell Theorem**, which applies for spherically symmetric mass or charge distributions).

This electric field $E$ is made up of 2 components : that $E_1$ due to the local charge on the disk and that $E_2$ due to the rest of the spherical shell : $E=E_1+E_2$.

The electric field due to the charge on the disk is $E_1=\frac{\sigma}{2\epsilon}$. It points both inwards and outwards. The electric field $E_2$ from the rest of the shell points only outwards, as from a point charge located at the centre of the sphere.

Inside the shell $E_1$and $E_2$ point in opposite directions, so the total electric field inside the shell is zero : $E_2+E_1=0$. This means that $E_2$ has the same magnitude as $E_1$, viz. $\frac{\sigma}{2\epsilon_0}$.

Outside the shell both components $E_1, E_2$ point in the same direction and add up to $E=\frac{\sigma}{\epsilon_0}$.

The force on the disk is equal to the charge on the disk $q_1$ times the electric field $E_2$ due to the rest of the spherical shell :

$$F_1=q_1 E_2=\sigma \pi r^2 \frac{\sigma}{2\epsilon_0}=\frac{1}{2\epsilon_0}(\frac{Q}{4\pi R^2})^2\pi r^2=\frac{Q^2r^2}{32\pi \epsilon_0 R^4}$$

The acceleration of the disk is $a_1=F_1/m$.

**Comment :**

You might wonder, where did we make use of the approximation that the disk is "negligibly small" compared with the sphere?

The expressions for $E, E_1, E_2$ are exact. They apply close to the surface of a uniformly charged spherical shell - ie when the "disk" is part of a sphere, like a bowl. If the disk is large but remains flat, as in your sketch, the electric field is much more difficult to calculate.

We used the approximation again when we calculated force $F_1$. We assumed the bowl was small and therefore approximately flat. Then the electric field is in the same direction across the bowl. However, for "large" bowls, although the magnitude of the electric field remains the same across them its direction varies. We then have to do a vector addition. Components of force parallel to the axis of the bowl add up, those perpendicular to the axis cancel out.

The expression for $F_1$ would be exact if we measured the radius of the disk/bowl not when it is flat, but across the rim of the curved bowl. This is equivalent to doing the "vector addition" of the forces across the surface of the bowl.

Then for the force between two hemispherical shells ($r=R$) we would get $F_1=Q^2/32\pi \epsilon_0 R^2$. (See https://www.youtube.com/watch?v=4EAcdEuunxI.)