This comes a bit late, but for any other people looking for a simple intuitive (non-mathematical) idea about PCA: one way to took at it is as follows:
if you have a straight line in 2D, let's say the line y = x. In order to figure out what's happening, you need to keep track of these two directions. However, if you draw it, you can see that actually, there isn't much happening in the direction 45 degrees pointing 'northwest' to 'southeast', and all the change happens in the direction perpendicular to that. This means you actually only need to keep track of one direction: along the line.
This is done by rotating your axes, so that you don't measure along x-direction and y-direction, but along combinations of them, call them x' and y'.
That is exactly encoded in the matrix transformation above: you can see it as a matrix transformation, but equivalently as a rotation of the direction in which you measure. Now I will refer you to maths literature, but do try to think of it as directions in which you measure.
PCA amounts to finding and interpreting a singular value decomposition (SVD) of $X$ (or any closely-related matrix obtained by various forms of "standardizing" the rows and/or columns of $X$). There are direct connections between any SVD of $X$ and any SVD of $XX^\prime$. I will exhibit the connections in both directions, by obtaining an SVD of one from an SVD of the other.
1. Any SVD of $X$ determines a unique SVD of $XX^\prime$.
By definition, an SVD of $X$ is a representation in the form
$$X = U\Sigma V^\prime$$
where $\Sigma$ is diagonal with non-negative entries and $U$ and $V$ are orthogonal matrices. Among other things, this implies $V^\prime V = \mathbb{I}$, the identity matrix.
Compute
$$XX^\prime = (U\Sigma V^\prime)(U\Sigma V^\prime)^\prime= (U\Sigma V^\prime)(V\Sigma U^\prime) = U\Sigma V^\prime V \Sigma U^\prime = U\Sigma \mathbb{I} \Sigma U^\prime = U (\Sigma^2) U^\prime.$$
Since $\Sigma^2$ is diagonal with non-negative entries and $U$ is orthogonal, this is an SVD of $XX^\prime$.
2. Any SVD of $XX^\prime$ gives (at least one) SVD of $X$.
Conversely, because $XX^\prime$ is symmetric and positive-definite, by means of an SVD or with the Spectral Theorem it can be diagonalized via an orthogonal transformation $U$ (for which both $UU^\prime$ and $U^\prime U$ are identity matrices):
$$XX^\prime = U\Lambda^2 U^\prime.$$
In this decomposition, $\Lambda$ is an $n\times n$ diagonal matrix ($n$ is the number of rows of $X$) with $r = \text{rank}(X)$ non-zero entries which--without any loss of generality--we may assume are the first $r$ entries $\Lambda_{ii}, i=1,2,\ldots, r$. Define $\Lambda^{-}$ to be the diagonal matrix with entries $1/\Lambda_{ii}, i=1,2,\ldots, r$ (and zeros otherwise). It is a generalized inverse of $\Lambda$. Let
$$Y = \Lambda^{-}U^\prime X$$
and compute
$$YY^\prime = \Lambda^{-} U^\prime XX^\prime U \Lambda^{-} = \Lambda^{-} \Lambda^2 \Lambda^{-} = \mathbb{I}_{r;n}.$$
(I have employed the notation $\mathbb{I}_{r;n}$ for an identity-like $n\times n$ matrix of rank $r$: it has $r$ ones along the diagonal and zeros everywhere else.) This result implies $Y$ must be a block matrix of the form
$$Y = \pmatrix{W^\prime & 0 \\ 0 & 0}$$
with $W$ an $r\times r$ orthogonal matrix. The zeros are matrices of appropriate dimensions to fill out the rest of $Y$ (which has the same dimensions as $X$). Let $p$ be the number of columns of $X$. Certainly $r \le p$. If $r \lt p$, we may extend $W$ to a $p\times p$ orthogonal matrix $V$. This can be done in many ways, but a simple one is to put $W^\prime$ in the upper left block, an identity matrix of dimensions $p-r\times p-r$ in the lower right block, and zeros everywhere else.
Since (by construction) $V^\prime V = \mathbb{I}_p$,
$$U^\prime X V = \Lambda Y V = \Lambda (V^\prime V) = \Lambda.$$
Upon left-multiplying by $U$ and right-multiplying by $V^\prime$ we obtain
$$U\Lambda V^\prime = (UU^\prime) X (VV^\prime) = X,$$
which is an SVD of $X$.
$$XX^\prime = (U\Sigma V^\prime)(U\Sigma V^\prime)^\prime= (U\Sigma V^\prime)(V\Sigma U^\prime) = U\Sigma V^\prime V \Sigma U^\prime = U\Sigma \mathbb{I} \Sigma U^\prime = U (\Sigma^2) U^\prime.$$
Since $\Sigma^2$ is diagonal with non-negative entries and $U$ is orthogonal, this is an SVD of $XX^\prime$.
Best Answer
The paper you cited (Donoho et al. 2013 Optimal Shrinkage of Eigenvalues in the Spiked Covariance Model) is an impressive piece of work which I confess I did not really study. Nevertheless, I believe that it is easy to see that an answer to your question is negative: using any kind of shrinkage estimator of the covariance matrix will not improve your PCA results and, specifically, will not lead to "better understanding of the structure in the data".
In a nutshell, this is because shrinkage estimators only affect the eigenvalues of the sample covariance matrix and not the eigenvectors.
Let me quote the beginning of the abstract of Donoho et al.:
The abstract goes on to describe paper's contributions, but what is important for us here is that the sample covariance matrix $S$ and its shrinked version $\hat\Sigma$ have the same eigenvectors. Principal components are given by projections of the data onto these eigenvectors; so they will not be affected by the shrinkage.
The only thing that can get affected are the estimates of how much variance is explained by each PC because these are given by the eigenvalues. (And as @Aksakal wrote in the comments, this can affect the number of retained PCs.) But the PCs themselves will not change.