Why do I keep losing my ‘text substitutions?’

Curl can be equated with the closed line integral in the limit that the encircled area $\Delta S$ goes to zero. However, we would have to do this in three components because curl is a vector. $$ (\nabla \times \vec{v})_x = \lim_{\Delta S \rightarrow 0} \frac{1}{\Delta S} \oint \vec{v}\cdot d\vec{l} $$ in the $yz$ plane and so on.

But what does it mean? Well it is easy to show that $$ \nabla \times \vec{v} = 2 \vec{ \omega} $$ As follows: $$ (\nabla \times \vec{v})_x = \partial_y v_z - \partial_z v_y = \partial_y (\vec{\omega} \times \vec{r})_z - \partial_z (\vec{\omega} \times \vec{r})_y $$ $$ (\nabla \times \vec{v})_x = \partial_y (\omega_x y - \omega_y x) - \partial_z ( \omega_z x - \omega_x z) = 2 \omega_x$$

and ditto for the other components $$ (\nabla \times \vec{v})_y = \partial_z v_x - \partial_x v_z = 2\omega_y$$ $$ (\nabla \times \vec{v})_z = \partial_x v_y- \partial_y v_x = 2\omega_z$$ i.e. the curl of a velocity field equals twice the angular velocity at that point. In other words it is angular velocity within a fluid flow that creates curl! You can imagine constructing a ``curl meter'' out of a little (infinitesimally small) paddle wheel which could be inserted into the fluid flow. If the paddle wheel turns then there is curl.

You've got a bunch of good answers now, let me add two things that I did not see:

The $\delta$-function curl: things can look twisty without being so.

Consider the field $$\vec V(x, y, z) = \frac{1}{x^2 + y^2}\begin{bmatrix}-y\\x\\0\end{bmatrix} = \hat\theta / r.$$Strangely, $\nabla\times\vec V=0$ at every point except the line $(x,y,z)=(0, 0, z)$ where it is not defined. This "looks" rotational: the $\hat\theta$ vector points around the origin. However, the exact scaling of $1/r$ is just what it needs for the curl to vanish.

So, any naive "the curl tells you how twisty something looks" interpretation is wrong, because here is a thing which looks twisty but has no curl.

Better: the force-torque interpretation.

Suppose $\vec V$ represents a force field -- in the usual physics sense of a force defined at every point in space, please, not the sci-fi sense of an invisible wall.

Now suppose we put a little pinwheel inside the force field and have it feel those forces. Without loss of generality, let's write it as a little circle of radius $\epsilon$ in the $xy$-plane about the origin, the points $\delta\vec r(\theta) = [\epsilon \cos\theta, \epsilon \sin\theta, 0]$ for $0\le\theta\lt 2\pi.$ The force on any point of this circle is approximately: $$\vec F(\theta) \approx \vec V(0, 0, 0) + \frac{\partial \vec V}{\partial x}~\epsilon~\cos(\theta) + \frac{\partial \vec V}{\partial y}~\epsilon~\sin(\theta).$$ To first order in $\epsilon$, of course, the net force on this pinwheel is just $\vec V(0, 0, 0)$: the terms linear in $\epsilon$ vanish.

The net torque, however, does not vanish: $$\delta \vec\tau = \int_0^{2\pi} d\theta~ \delta\vec r(\theta)\times\vec F(\theta) \approx \epsilon \int_0^{2\pi} d\theta~\hat z~\big(\partial_x V_y \cos^2\theta - \partial_y V_x \sin^2\theta \big),$$with all the other terms disappearing as we integrate plain sines and cosines or occasionally $\sin\theta\cos\theta = \frac12 \sin(2\theta),$ all of which oscillate about $0$. Only $\cos^2$ and $\sin^2$ oscillate about a different average, namely $\frac12$. Performing the integral therefore leads to the small torque:$$\delta \vec\tau \approx \frac12~\epsilon ~\hat z \left(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\right).$$This is $\vec \tau = \epsilon~\hat z~(\hat z \cdot (\nabla \times \vec V)),$ so the proper extension of this to all coordinates is simply: at position $\vec r$ insert a pinwheel in the $\hat n$ direction, the torque per unit radius on that pinwheel is simply $\hat n \cdot (\nabla \times \vec V).$

So the curl of a force field is the torque-per-unit-radius on a small pinwheel, pointed in the direction of greatest torque.

Since a velocity field at small distances will generally influence particles by a linear drag $\vec F \propto \vec v$, this is also a good interpretation for velocity fields: insert a little pinwheel that you are holding fixed, and the curl tells you the torque that the fluid will start exerting on that pinwheel (which you will have to oppose to keep it fixed).

The $\hat \theta / r$ field, therefore, does not rotate a small object as it flows past it. You can roughly understand this by describing the pinwheel not as a little circle of radius $\epsilon$ but a little trapezoid $\delta r, \delta \theta.$ The fluid stays in contact with the outer surface for a length $\delta \theta~(r + \delta r)$ but is only in contact with the inner surface for the length $\delta \theta~ r.$ By making the force go like $\vec F = U_0~\hat \theta/r,$ the work done on the inner surface travelling around the loop is $-U_0~\delta\theta~r/r ~+~ U_0 \delta\theta (r + \delta r) / (r + \delta r) = 0,$ as it must be if the pinwheel doesn't want to rotate that way.