Edit: Thank you, John, for improving my newbie formatting. I hope there is a way to bold the vectors in math mode; I haven't found a way to do it yet.
I think I can provide a much clearer (and shorter) explanation of singular vectors versus eigenvectors.
First, I encourage you to see an $(m \times n)$ real-valued matrix $A$ as a bilinear operator between two spaces; intuitively, one space lies to the left ($R^m$) and the other ($R^n$) to the right of $A$. "Bilinear" simply means that $A$ is linear in both directions (left to right or right to left). The operations $A$ can perform are limited to scaling, rotation, and reflection, and combinations of these; any other kind of operation is non-linear.
$A$ transforms vectors between the two spaces via multiplication:
$x^T$ A = $y^T$ transforms left vector $x$ to right vector $y$.
$x = A y$ transforms right vector $y$ to left vector $x$.
The point of decompositions of $A$ is to identify, or highlight, aspects of the action of $A$ as an operator. The eigendecomposition of $A$ clarifies what $A$ does by finding the eigenvalues and eigenvectors that satisfy the constraint
$A x = \lambda x$.
This constraint identifies vectors (directions) that $A$ does not rotate, and the scalars $\lambda$ associated with each of those directions.
The problem with eigendecomposition is that when the left and right space are different spaces, there really isn't a sense in which $A$'s action can be described as involving a "rotation", because the left and right spaces are totally separate, not "oriented" relative to one another. There just isn't a way to generalize the notion of an eigendecomposition to a non-square matrix $A$.
Singular vectors provide a different way to identify vectors for which the action of $A$ is simple; one that does generalize to the case where the left and right spaces are different. A corresponding pair of singular vectors have a scalar $\sigma$ for which $A$ scales by the same amount, whether transforming from the left space to the right space or vice-versa:
$ x^T A = \sigma y^T$
$\sigma x = A y$.
Thus, eigendecomposition represents $A$ in terms of how it scales vectors it doesn't rotate, while singular value decomposition represents $A$ in terms of corresponding vectors that are scaled the same, whether moving from the left to the right space or vice-versa. When the left and right space are the same (i.e. when $A$ is square), singular value decomposition represents $A$ in terms of how it rotates and reflects vectors that $A$ and $A^T$ scale by the same amount.
My original "answer" is below, however I hope you will agree that the one above is far superior.
Maybe it would help to have an intuitive (by which I mean a geometrical) explanation of what singular vectors and eigenvectors are.
A singular value $s$ of a square ($n \times n$) matrix $A$ is a nonzero scalar for which there exist unit vectors $x$ and $y$ that satisfy the two equations:
$A^T x = \sigma y$ and
$A y = \sigma x$.
$x$ and $y$ are the left and right singular vectors, respectively, of $A$. A geometric interpretation of this is as follows. Let $u(t) = t x$ and $v(t) = t y$
be the parametric equations defining the lines in the direction of the vectors $x$ and $y$, respectively. Then
$A^T u(t) = A^T tx = t A^T x = t \sigma y = v(\sigma t)$,
$A v(t) = A ty = t A y = t \sigma x = u(\sigma t)$.
In words, the bilinear operator $A$ does the following: $A^T$ maps the line $u(t)$ to the line $v(\sigma t)$, where the change in the parameter ($t$ to $\sigma t$) means "your speed along the line $v$ is multiplied by $\sigma$". Likewise, $A$ maps the line $v(t)$ to the line $u(\sigma t)$.
The eigenvectors of $A$ have a different interpretation: given any vector $x$,
(1) project $x$ onto the eigenvectors of $A$ (i.e. represent $x$ in the set of basis vectors that are the eigenvectors of $A$).
(2) scale each projected component of $x$ by the corresponding eigenvector; the resulting vector $y = A x$ (but $y$ is represented in the eigenvector basis).
(3) "un-project" the vector $y$ back to the original coordinate system.
Thus, the eigenvectors define a new set of basis vectors along which the scaling occurs; the singular vectors define lines that $A$ maps, one to the other.
I hope that's the kind of intuitive answer you were looking for. It certainly was the kind I was looking for.