Why the work done in a conservative field around a closed circle does not vanish when calculated in cylindrical coordinates?

The problematic line in your reasoning is in assuming that $$\nabla \times \vec F=0$$ implies that $$\vec F$$ is conservative, i.e. that $$\vec F = \nabla \phi$$ for some $$\phi$$. If this were true, then it would imply that $$\oint \vec F \cdot \mathrm d\vec r = 0$$ via the fundamental theorem of vector calculus, but alas it is not.

We are only permitted to make the leap $$\nabla \times \vec F = 0 \implies \vec F = \nabla\phi$$ if $$\vec F$$ is differentiable on a simply-connected domain. Simple-connectedness means that every loop in the domain can be continuously shrunk to a point without leaving the domain. This is not the case here - $$\vec F$$ is not differentiable (or even defined) anywhere on the $$z$$-axis, so any loop encircling it cannot be shrunk to a point. On such domains, $$\nabla \times \vec F = 0$$ is necessary but not sufficient for $$\vec F$$ to be conservative.

An alternate viewpoint is provided by the given solution. If you try to find some $$\phi$$ such that $$F = \nabla \phi$$, you will only be able to do so on a domain which does not completely surround the origin. For example, $$\phi(x,y,z) = \tan^{-1}(y,x)$$ is defined for all of $$\mathbb{R^3}$$ except for any arbitrary slice (across which the angle jumps by $$2\pi$$), but that's not good enough, as $$\vec F=\nabla \phi$$ has to hold everywhere in order for $$\vec F$$ to be conservative.

The core idea in this problem is that integrating the curl over the region inside the unit circle is not a valid operation, because the curl is not defined at the origin.*

You can integrate over a region that looks like the unit circle, but has a wedge cut out of it so that the boundary goes most of the way along the unit circle, comes in along an almost-radial line, loops around the origin, and then goes back to the circle along another almost-radial line. The limit of this wedge being infinitesimally small is the "cut line" in the quote.

(Another approach to removing the origin is to have an infinitesimally-small circle around the origin, in which case you end up getting a curl integral over "the unit circle minus a point at the origin". This integral corresponds to the work done by going around the unit circle, minus the work required to go around the inner circle.)

* As pointed out by @NoLongerBreathedIn, it is possible to use a generalization of integrating over functions (integrating over distributions) to accommodate singularities like the one at the origin. This approach seems to be outside the scope of the question, which asks why Arken and Weber assert that they need to remove a "cut line" from the circle to invoke Stokes' theorem in the context of function integrals.