I believe myself to have an intuition of the basic vector differential operators, $\mathrm{grad} \equiv \nabla$, $\mathrm{div} \equiv \nabla \cdot$, $\mathrm{curl} \equiv \nabla\times$ and the Laplace operator which is defined as $\nabla^2 \equiv \nabla \cdot \nabla$.
However, the double curl operator which I encountered in the electromagnetic wave equation derived from Ampére's and Faraday's laws has always been very abstract to me. $$\nabla\times\nabla\times \equiv \nabla(\nabla \cdot) - \nabla^2$$
Any intuition would be helpful!
Best regards.
Multivariate Fourier analysis tells us that in principle one can generally decompose any sufficiently nice vector-valued function $\vec F(\vec r)$ as a superposition of elementary plane waves:
$\vec F(\vec r)=\int\int \int_{\vec k} \vec A(\vec k) e^{ i \vec k \cdot \vec r } \ d\vec k$. Thus it is useful to understand how various vector calculus operators treat each elementary term in this superposition. Perhaps this understanding qualifies as some form of "computational intuition" about how these operators behave.
Notation for an elementary plane wave.
For fixed values of $\vec k$ and $\vec A(\vec k)$ we write the elementary plane wave as
$F= f \vec A$ where $f=e^{ i (\vec k \cdot r) }$ is a scalar function of the position vector $\vec r$. Note that only $f$ (not $\vec A$) depends on position. That simplifies all that follows.
The product rules for divergence, curl, etc.
One can check that since $\vec A$ is constant in space,
$grad f= i \vec k f$
$div (F)= div(f\vec A)= grad f \cdot \vec A =i (\vec k \cdot \vec A) f$
$curl (F)= curl (f \vec A) = grad f \times \vec A= ( i)( \vec k\times \vec A) f$
The fundamental decomposition formula for the Laplacian of a Vector Field
so (1) $curl(curl f\vec A)= - \vec k\times (\vec k\times \vec A) f$
Similarly
(2) $\nabla ^2 (f \vec A)= -(\vec k\cdot \vec k) \vec A f$
Next $\nabla (\nabla \cdot F)=grad (div (f\vec A) =grad (grad f\cdot \vec A)= grad (i f \vec k \cdot \vec A)$
$ = (i)^2 f (\vec k) (\vec k \cdot \vec A)$; that is,
(3) $\nabla (\nabla \cdot F) = -(\vec k) [(\vec k \cdot \vec A)] f$.
Comparing (1)+(3) vs (2), let's see what happens when we hold $\vec k$ fixed and when $\vec A$ is decomposed into components that are either parallel or perpendicular to $\vec k$. Note that (1),(2) and (3) depend linearly on $\vec A$. One checks easily that each such component of $\vec A$ satisfies the desired relation (1)+(3) =(2).
P.S. For simplicity, you can WLOG assume that $\vec k=(0,0,1)$ and then use linearity to justify checking the 3 purest elementary cases $\vec A= (1,0,0), (0,1,0),$ and $ (0,0,1)$.
P.P.S Note that when $\vec A \cdot \vec k=0$ the divergence of $F$ is zero, and when $\vec A || \vec k$ then the curl of $F$ is zero. This reveals how a general such $F$ can be decomposed into curl-free and divergence -free parts. (Helmholtz decomposition).