Where does the centripetal force come from? How to calculate?

A particle at the surface of the water experiences two forces. Once is the force of gravity which acts vertically downwards. The other is the buoyancy force from the surrounding water, which acts perpendicular to the surface of the water. If the buoyancy force is $B$ and the water surface is at an angle $theta$ to the horizontal then we can resolve forces horizontally and vertically to get

$F_h = B sin theta F_v = B cos theta - mg$

The net vertical force must be zero, so we have

$B cos theta = mg Rightarrow F_h = mg tan theta$

The net horizontal force is the centripetal force. This comes from the horizontal component of the buoyancy force. If we assume all particles at the surface of the water are travelling in horizontal circles then the centripetal force will depend on the mass $m$ of the particle, the radius $r$ of the horizontal circle in which the particle is travelling and the angule velocity $omega$. So

$mgtan theta = mromega^2 Rightarrow tan theta = frac{romega^2}{g}$

If we assume all particles have the same angular velocity the $tan theta$ is proportional to $r$. For particles near the centre of the bucket $r$ and $theta$ are small. As you move further away from the centre of the bucket $r$ increases and so $theta$ must also increase.

See, a free surface of a fluid (here a liquid) is a surface of constant pressure. This is so because fluids cannot bear any tangential stress. According to your question, I'll assume the fluid tube rotates about a vertical axis and does not translate. I'll define the co-ordinate axes as the following and take two points on the surface, A and B.

As I have mentioned, the pressure at both A and B must be the same, as both lie on the fluid surface.

Note : This is only a 2D view of the surface, a 3-dimensional argument will give a paraboloid shape.

Let this have a density of $rho$. Now, you should note that in a rotating fluid, an excess pressure is developed in the direction perpendicular to the angular velocity, $omega$. This pressure mainly comes from the thrust from neighbouring molecules. One can easily reason that a molecule by itself would follow a tangential path had not the surrounding molecules constrained it to move in a circular path. This creates a $∆P$. To see how, consider an element of length $dx$ along a cylindrical part of the rotating fluid, which is at a distance $x$ from the other end (the centre of rotation).

The net force on this element is $$F = SdP$$ This force is the centripetal force which rotates this element along a radius $x$ Hence Integrating the expressions, $$int_0^P dP$$ = $$int_0^x rhoomega^2 xdx$$ Thus $$P = frac{rhoomega^2 x^2}{2}$$ Now, going back to the original problem, as in the figure a horizontal distance of $x$ gives an excess pressure of $$frac{rhoomega^2 x^2}{2}$$ while an vertical elevation of $y$ decreases pressure by $yrho g$. Equating these two, $$ y = frac{omega^2 x^2}{2g}$$ which is the equation of a parabola.