$\nabla\times\vec{A}$ is the solenoidal component of the vector field: it is the divergenceless component.
A good way to intuitively visualise the Helmholtz theorem is to think in Fourier space, so that all fields become their Fourier transforms. In this visualisation, $\nabla\times \vec{E}$ is the component of the Fourier transform that is at right angles to the wavevector. Then the differential operations become dot and cross products with the position in Fourier space i.e. the wavevector of plane wave component. So, instead of $\vec{E}(\vec{r})$, we have now a field $\tilde{\vec{E}}(\vec{k})$ specifying the superposition weight (amplitude) of a plane wave component $\exp(i\,\vec{k}\cdot\,\vec{r})$ in the Fourier decomposition of the field. The Fourier transforms of the following $\nabla$ operations on vector and scalar fields $\vec{E}(\vec{r})$ and $\phi(\vec{r})$ are:
$$\nabla\phi(\vec{r}) \leftrightarrow i\,\vec{k}\,\tilde{\phi}(\vec{k})$$
$$\nabla\cdot\vec{E}(\vec{r}) \leftrightarrow i\,\vec{k}\,\cdot\tilde{\vec{E}}(\vec{k})$$
$$\nabla\times\vec{E}(\vec{r}) \leftrightarrow i\,\vec{k}\,\times\tilde{\vec{E}}(\vec{k})$$
where the tilde quantities are the Fourier transforms. So, at each point in space, $\tilde{\vec{E}}(\vec{k})$ resolve the vector into unique components parallel to and orthogonal to the wavevector (the position vector in Fourier space):
$$\begin{array}{lcl}\tilde{\vec{E}}(\vec{k}) &=& \tilde{E}_\parallel(\vec{k})\,\frac{\vec{k}}{|\vec{k}|}+\tilde{E}_\perp(\vec{k})\,\frac{\vec{k}\times \tilde{\vec{E}}(\vec{k})}{|\vec{k}|\,|\tilde{\vec{E}}(\vec{k})|}\\&=&-i\,\left(\mathfrak{F}(\nabla.\vec{E}) \,\hat{k} + \mathfrak{F}(\nabla\times\vec{E})\right)\end{array}$$
where $\mathfrak{F}$ is the Fourier transform. The curl is the component of the field at right angles to the unit radius $\hat{k}$ (modulo the scaling constant $i$), since $\hat{k}\cdot(\vec{k}\times \tilde{\vec{E}})=0$ and the divergence is the component along the unit radius $\hat{k}$