When you're using a norm-conserving pseudopotential, the charge density $\rho^\mathrm{NC}(r)$ is calculated directly from the modulus-squared of the wavefunction:
\begin{align}\tag{1}
\rho^\mathrm{NC}(r) = \sum_{bk}\left\vert\psi_{bk}(r)\right\vert^2,
\end{align}
where $b$ and $k$ are band and k-point indices, respectively, and I've restricted the summation to over occupied bands, for simplicity (in general, we'd also need to include the band occupation). By convention, we also omit the fundamental charge of an electron, $e$, since it has unit value in atomic units.
In order to fully represent this charge density, we need to have a Fourier grid which is twice as big as the wavefunction cut-off wavevector in each direction. This factor of two comes directly from the fact that the density is the wavefunction squared.
In an ultrasoft (or PAW) method, the charge density is no longer simply the modulus-squared of the Bloch functions, but gains an "augmentation" charge term,
\begin{align}\tag{2}
\rho^\mathrm{USP}(r) = \sum_{bk}\left(\left\vert\psi_{bk}(r)\right\vert^2 + \sum_{nmI}Q^I_{nm}(r)\langle\psi_{bk}\vert\beta_{n}^I\rangle\langle\beta_{m}^I\vert\psi_{bk}\rangle\right),
\end{align}
where $I$ is an atom index, and $n$ and $m$ are projector indices. $Q^I_{nm}(r)$ are the augmentation charges themselves.
If we have a Fourier grid twice as large as the cut-off wavevector, then we can represent the first term exactly; however, the second term depends on $Q^I_{nm}(r)$, which is not related to $\psi$. Can $Q^I_{nm}(r)$ be represented exactly on this Fourier grid, or does it need a larger grid?
$Q(r)$ is actually designed to have larger Fourier components, this is in fact why the ultrasoft/PAW method can give lower cut-off energies for $d-$ and $f-$block elements. This means that it cannot usually be represented well on the same Fourier grid we used for the soft charge density, and we need a larger grid for the full augmented charge density. It is the size of this "augmentation charge grid" which is controlled by EAUG.
