How does the cokernel of the J-homomorphism count exotic spheres? The wikipedia article on the J-homomorphism says that "the cokernel of the J-homomorphism is of interest for counting exotic spheres".  I'd like to think this makes some sort of philosophical sense; as I understand it, the homomorphism comes from a Hopf construction, which isn't really a smooth sort of thing, but on the other hand it still feels like constructions using the (stable) (special) orthogonal group should somehow keep us in the non-exotic world.  Is this at all close to the right intuition?
 A: This isn't really different from the other answer, but I love this story so I wanted to try telling it a little slower.  (I will also completely elide all technical details.)
Pontryagin-Thom says that any element of any stable homotopy group of spheres is represented by a stably framed manifold.  A natural question to ask (and the first question I asked myself when I learned about P-T) is how simple you can make the representing manifold?  The simplest possible manifolds are standard spheres with some stable framing.  Relative to the standard  with the standard stable framing (i.e. coming from the embedding of $S^n$ in $\mathbb{R}^{n+1}$ with the outward going normal), you've chosen some map from the appropriately stabilized trivial bundle to itself, i.e. an element of SO(n+k) for every point on your sphere.  That is these are classified by elements of $\pi_n(SO)$.  This gives you exactly the J-homomorphism.  So the image of J is the part of the homotopy groups of spheres which comes from standard spheres.
Ok, but what about more complicated manifolds?  Well in 2-dimensions you already see roughly how to do this: you start with a surface, you try to find a circle whose induced framing is such that you can cut along it and get a smaller genus surface.  However, the Arf invariant on homology gives an obstruction to being able to do this, so you end up with either a trivial manifold or the torus with a particular stable framing (the Lie group framing).  The nontrivial element of the second stable stem comes from this torus.  We want to mimic this in higher dimension, and this is where surgery theory enters.  That is we want to do surgery to decrease the complexity of our manifold, but we have to be a bit careful because we're doing stably framed surgery and so we have to make sure we have choose our elements in homology so that we can glue in stably framed discs.  The key results from surgery theory are then that you can successively kill off all homology except possibly a single class in middle dimension.  This is the Kervaire problem: are there manifolds of a given dimension with no homology except for one class in middle degree which you can't do framed surgery along?  (Answer: exactly one such manifold in dimensions 2, 6, 14, 30, 62, and no such manifolds in any other dimension except possibly 126 which is still open.  The explicit manifolds are known in dimensions 2 (flat torus), 6, 14, and 30, but not dimension 62.)
Ok, so outside the elements corresponding to the Kervaire manifolds we've done surgery to get the manifold down to a smooth manifold which is topologically a sphere.  But it doesn't have to have the standard smooth structure!  So such classes correspond to exotic spheres together with a stable framing.  Now you need to know how many stable framings you'd expect an exotic sphere to have.  First you want to show they all have some stable framing by immersing them into a large Euclidean space, and then second you want to argue that again the changes of framings are just given by composing with elements in the image of J.
All told: up to stably framed cobordism every manifold comes from a standard sphere with an interesting framing connect-sum your favorite framing on some exotic sphere connect-sum one of five or six explicit Kervaire manifolds generalizing the flat torus.
A: The connection is very concrete.
Namely let us use the standard notations: $\theta_n$ for the group of homotopy $n$-spheres, and $bP^{n+1}$ for the subgroups formed by those homotopy spheres which bound a parallelizable $n+1$-dimensional manifold. If $n$ is odd, then the quotient group
$\theta_n/bP^{n+1}$ is isomorphic to the cokernel of the stable $J$ homomorphism
$J: \pi_n(SO) \to \pi^s(n).$
The proof is, that any framed $n$-manifold representing (by the Pontrjagin construction) an element in the stable homotopy group of spheres can be made by by framed surgery  $(n-1)$-connected, i.e. a homotopy sphere. So any element of $\pi^s(n)$ can be represented by a homotopy sphere. It is not hard to see, that the represented element in $\pi^s(n)$ is trivial iff the homotopy sphere lies in $bP^{n+1}.$
For n even the above described surgery can be performed only iff the Kervaire invariant of the framed manifold is zero. Hence in this case the quotient $\theta_n/bP^{n+1}$ will be isomorphic to the cokernel of the $J$ homomorphism in the kernel of the Kervaire invariant.
