The distance should be easy to determine in closed-form from $r$ (cylinder radius), $\overline{C}$ (a vector along the cylinder's center) and $\overline{P}$ (a vector representing a ray from that centerline to the point $p$).
My first stab...
$$ d = r \tan\left(\arccos{\left(\! \frac{\overline{C}\cdot\overline{P}}{|\overline{C}| |\overline{P}|} \!\right)}\right) $$
I think a more fundamental way to approach the problem is by discussing geodesic curves on the surface you call home. Remember that the geodesic equation, while equivalent to the Euler-Lagrange equation, can be derived simply by considering differentials, not extremes of integrals. The geodesic equation emerges exactly by finding the acceleration, and hence force by Newton's laws, in generalized coordinates.
See the Schaum's guide Lagrangian Dynamics by Dare A. Wells Ch. 3, or Vector and Tensor Analysis by Borisenko and Tarapov problem 10 on P. 181
So, by setting the force equal to zero, one finds that the path is the solution to the geodesic equation. So, if we define a straight line to be the one that a particle takes when no forces are on it, or better yet that an object with no forces on it takes the quickest, and hence shortest route between two points, then walla, the shortest distance between two points is the geodesic; in Euclidean space, a straight line as we know it.
In fact, on P. 51 Borisenko and Tarapov show that if the force is everywhere tangent to the curve of travel, then the particle will travel in a straight line as well. Again, even if there is a force on it, as long as the force does not have a component perpendicular to the path, a particle will travel in a straight line between two points.
Also, as far as intuition goes, this is also the path of least work.
So, if you agree with the definition of a derivative in a given metric, then you can find the geodesic curves between points. If you define derivatives differently, and hence coordinate transformations differently, then it's a whole other story.
Best Answer
Let me try to help with expressing your problem. The following is related:
Find minimum distance between two points on a cylinder, one on the top flat surface $A$ (not its circle center) and another point $B$ on its curved surface.
EDIT1:
$$ AO=a\,;\angle AOQ =\alpha; AOP =\theta; \, QB= z ; $$
The problem is best formulated using development of the cylinder curved surface.
Given $(r,\alpha, a,z) $ you want to minimize total length
$$ AP +PB = \sqrt{a^2+r^2-2 a r \cos \theta} +\sqrt{z^2+r^2( \alpha-\theta)^2 }\tag1$$
This should be differentiated w.r.t $\theta$ and set to zero leading to :
$$ \frac {a \sin \theta}{r (\alpha-\theta)} = \sqrt{\frac {a^2+r^2-2 a r \cos \theta}{z^2 +{r^2( \alpha-\theta)^2 }} }\tag2$$
which should be solved. A numerical solution can be attempted as it is a transcendental equation having no closed form solution.
A minimum is seen somewhere in the middle $\theta \approx 0.86$ angle in domain for values assumed $(a,r,\alpha,z)=(.5,1, \pi/2,1.25 ) $ where $ \theta $ is on x-axis and total length of two line segments $ APB$ on y-axis.
EDIT2:
It is instructive here to mention geodesics in 3D. Label the vertex opposite to $Q$ as $C,$ to make $ PQBC $ a rectangle. Now the point $P$ should move such that line $APB$ should be straight for minimum length, or in other words
$$\boxed{ \angle APO= \angle BPC = \psi } \tag3 $$
$$ \sin \psi =\frac{a \sin \theta}{\sqrt{ z^2 +r^2( \alpha-\theta)^2 }} =\frac{r( \alpha-\theta) } {\sqrt {z^2 + r^2( \alpha-\theta)^2 } } \tag4 $$
Minimum distance from axis to line on flat top surface is the Clairaut constant for surfaces of revolution. $$ = a \sin \psi \tag5 $$
It is thus seen that the surface of revolution can have abrupt change in slope needing no smoothness and continuity.