What is the importance of eigenvalues/eigenvectors?
Short Answer
Eigenvectors make understanding linear transformations easy. They are the "axes" (directions) along which a linear transformation acts simply by "stretching/compressing" and/or "flipping"; eigenvalues give you the factors by which this compression occurs.
The more directions you have along which you understand the behavior of a linear transformation, the easier it is to understand the linear transformation; so you want to have as many linearly independent eigenvectors as possible associated to a single linear transformation.
Slightly Longer Answer
There are a lot of problems that can be modeled with linear transformations, and the eigenvectors give very simply solutions. For example, consider the system of linear differential equations \begin{align*} \frac{dx}{dt} &= ax + by\\\ \frac{dy}{dt} &= cx + dy. \end{align*} This kind of system arises when you describe, for example, the growth of population of two species that affect one another. For example, you might have that species $x$ is a predator on species $y$; the more $x$ you have, the fewer $y$ will be around to reproduce; but the fewer $y$ that are around, the less food there is for $x$, so fewer $x$s will reproduce; but then fewer $x$s are around so that takes pressure off $y$, which increases; but then there is more food for $x$, so $x$ increases; and so on and so forth. It also arises when you have certain physical phenomena, such a particle on a moving fluid, where the velocity vector depends on the position along the fluid.
Solving this system directly is complicated. But suppose that you could do a change of variable so that instead of working with $x$ and $y$, you could work with $z$ and $w$ (which depend linearly on $x$ and also $y$; that is, $z=\alpha x+\beta y$ for some constants $\alpha$ and $\beta$, and $w=\gamma x + \delta y$, for some constants $\gamma$ and $\delta$) and the system transformed into something like \begin{align*} \frac{dz}{dt} &= \kappa z\\\ \frac{dw}{dt} &= \lambda w \end{align*} that is, you can "decouple" the system, so that now you are dealing with two independent functions. Then solving this problem becomes rather easy: $z=Ae^{\kappa t}$, and $w=Be^{\lambda t}$. Then you can use the formulas for $z$ and $w$ to find expressions for $x$ and $y$..
Can this be done? Well, it amounts precisely to finding two linearly independent eigenvectors for the matrix $\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$! $z$ and $w$ correspond to the eigenvectors, and $\kappa$ and $\lambda$ to the eigenvalues. By taking an expression that "mixes" $x$ and $y$, and "decoupling it" into one that acts independently on two different functions, the problem becomes a lot easier.
That is the essence of what one hopes to do with the eigenvectors and eigenvalues: "decouple" the ways in which the linear transformation acts into a number of independent actions along separate "directions", that can be dealt with independently. A lot of problems come down to figuring out these "lines of independent action", and understanding them can really help you figure out what the matrix/linear transformation is "really" doing.
A short explanation
Consider a matrix $A$, for an example one representing a physical transformation (e.g rotation). When this matrix is used to transform a given vector $x$ the result is $y = A x$.
Now an interesting question is
Are there any vectors $x$ which does not change their direction under this transformation, but allow the vector magnitude to vary by scalar $ \lambda $?
Such a question is of the form $$A x = \lambda x $$
So, such special $x$ are called eigenvector(s) and the change in magnitude depends on the eigenvalue $ \lambda $.
The behaviour of a linear transformation can be obscured by the choice of basis. For some transformations, this behaviour can be made clear by choosing a basis of eigenvectors: the linear transformation is then a (non-uniform in general) scaling along the directions of the eigenvectors. The eigenvalues are the scale factors.