This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same time. The potential energy of this force is always greater than the excitation energy of the composite system, so if you force the bosons to sit at the same point, you will excite one of them, so that the composing fermions are no longer in the same state, and the two particles become distinguishable. The scale for this effective repulsion is the decay-length of the wavefunction of the composing fermions, and this repulsion is what leads matter to feel hard.
The reason you haven't heard this is somewhat political--- there are people who say that the exclusion principle is not the cause of the repulsive contact forces in ordinary matter, that this force is electrostatic, and despite this being ridiculously false, nobody wants to get into the mud and argue with them. So people don't explain the fermionic exclusion principle forces properly.
If you have a two-fermion composite which is net bosonic, like a H atom with a proton nucleus and spin-polarized electron, when you bring the H-atoms close, the energy of the electronic ground state is the effective Hamiltonian potential energy for the nuclei. When the nuclei are close enough so that the electronic wavefunctions have appreciable overlap, you get a strong repulsion. You can see that this repulsion is pure Pauli, because if the electrons have opposite spins, you don't get repulsion at short distances, you get attraction, and the result is that you form an H2 molecule of the two H atoms.
You can see this exclusion force emerge in an exactly solvable toy model. Consider a 1d line with two attractive unit delta function pontetials at positions a and -a, each with a fermion attached in the ground state. Each one has an independent ground state wavefunction that has the shape $exp(-|x|)$, but when the two are together at separation 2a, the two states are deformed, and the ground state energy for the fermions goes up. The effect is quadratic in the separation, because the ground state (one fermion) goes down in energy, and the first excited state goes up in energy, and to leading order in perturbations, the two are cancelling when both states are occupied. To next leading order, the effect is positive potential energy, a repulsion. This potential is the effective potnetial of the two delta functions when you make them dynamical instead of fixed.
The maximum value of the repulsive potential in this model is exactly where the model breaks down, which is at a=1. At this point, the ground state is exp(-2x) to the left of -1, constat between the two delta functions, then exp(2x) to the right, with energy -2, and the first excited state is constant to the left of -1, a straight line from -1 to 1, and constant past 1, with energy 0. The result is a net energy of -1 unit. This is half the binding energy of the two separated delta functions, which is -2.
This effect is the exclusion repulsion, and it reconciles the fermionic substructure with the net bosonic behavior of the particle. You can only see the substructure when the wavefunction of the boson is concentrated enough to have appreciable overlap on the scale of the composing fermion wavefunctions, and this is why you need high energies to probe the compositeness of the Higgs (or for that matter, the alpha particle). To get the wavefunctions to sit at the same point to this accuracy, you need to localize them at high energy.
Er ... nothing prevents this. That's what a Bose-Einstein condensate is: lots of bosons in the same place and quantum state.
You are observing that the state is not perfectly localized, but that is a consequence of the state not being exactly zero momentum. Ultimately the Heisenberg principle puts a lower limit on how localized they could be.
If the bosons are composite objects (like Helium atoms, say) then you can write the state in terms of their constituent parts and the Fermionic bits have to obey the Pauli principle.
Best Answer
Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities, $$\text{HH}, \text{HT}, \text{TH}, \text{TT}.$$ There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $\text{HH}$ and $\text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $\text{HT}$ and $\text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities, $$\text{two heads}, \text{two tails}, \text{one each}$$ and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.