To clarify the nonrigorous physics version of the argument--- the "overlap" of vacua at one point independent of other points only fully makes sense if the vacuum is a product state over separate points, i.e. if the field theory is ultra-local, meaning it has no derivative terms. But the principle is exactly the same when you have derivative terms, as any local change in the vacuum is amplified by translation invariance to a zero inner product with the original vacuum.
The same argument works in momentum space to establish that two free theories with different mass have orthogonal vacua in infinite volume. You can work out the inner product of the Gaussian states with each other at each k, and the inner product is the product over all k of these inner product. Knowing that there is some k region where the inner product is less than 1, and knowing that in the infinite volume limit, the number of k-variables in any region diverges, you conclude that the inner product, which is the product over all k of the inner product at k, goes to zero, and there's nothing more to it.
But in this example, you got an inequivalent vacuum because you changed the mass. The Haag thing is talking about turning on an interaction which is renormalized so as to keep the vacuum energy and the mass of the single particle excitations exactly constant as you turn it on. The Haag theorem then shows that the inner product is still zero, for the same trivial reason--- there is a local difference in the vacuum, and whatever the local difference is, it multiplies out to zero inner product due to infinite volume.
It helps for the purpose of illustration to state the Haag principle in a path integral, rather than in a rigorous formalism. The inner product of two vacua with two different values of the coupling (on a lattice, so that there are no renormalization issues), may be written down as a path integral in the sudden approximation: you have a boundary at t=0 where $\lambda$ has one value on one side, and another value on the other. The question "What is the probability amplitude for the vacuum to remain a vacuum under this sudden perturbation?" is the value of the path integral with vacuum boundary conditions on either side.
The obvious answer, whether the perturbation is sudden, or soft, whether it is for a strong interaction or weak, is always zero, because there is some nonzero probability per unit spatial volume of creating excitations from any perturbation, which means that the probability of having no excitations at the end of the perturbation is zero, so there is zero overlap. Again, it is only a question of infinite volume, and there is nothing extraordinary about it. The only notable thing about it is that it is possible to make it rigorous relatively easily given the methods of the 1960s.
Arnold Neumaier answered question 1 completely, and gave a specific model in a reference for question 2, but for question 2, you wanted a specific example model, and here's how you get all of them.
The key point is that you don't have a Haag theorem exactly when there is no particle production in the sudden approximation, which is tantamount to saying that the interaction cann't produce particles in vacuum. Then the sudden approximation does nothing at all, per unit area, or over the whole space, and the vacuum is unchanged. If there is no particle creation/annihilation from vacuum due to the interaction in the nonrelativistic theory, there is no particle annihilation into vacuum either (by Hermiticity of H) and there are no vacuum bubbles, the interacting vacuum is exactly the same as the noninteracting vacuum.
A simple example of a local nonrelativistic field theory with no vacuum bubbles, and still with local interactions (in whatever spatial dimensions), is
$$S = \int \bar\psi ( i\partial_t - {\nabla^2 \over 2m})\psi + \lambda \bar\psi \bar\psi \psi\psi $$.
This is the case of conserved particle number and delta-function interactions. It makes sense without any coupling renormalization in 1+1 d, but in 2+1 d it requires logarithmic renormalization to take a continuum limit, and in higher dimensions, it requires a power-law renormalization, but in any case, consider this on a lattice and in infinite volume. This interactions preserves the particle number N in all states, so it never produces particles starting in a vacuum.
Another example, this time which does not conserve N, so that there is particle creation and annihilation in some states, is the two species model (i ranges from 1 to 2)
$$ S = \int \sum_i \bar\psi_i (i\partial_t - {\nabla^2\over 2m})\psi_i - \sum_i \bar\psi_i\bar\psi_i\psi_i\psi_i -g( \bar\psi_1 \psi^2 \psi^1 + \bar\psi_1 \bar\psi_2\psi_1)$$
The key point is that the creation of type 2 particles always requires a type 1 particle to seed it, and the type 1 particle is conserved. This is what's going on in the Lee model, where N type particles can create C type particles, but the number of N type particles isn't modified in the process. Again, even though the type 1 particles create type 2 particles, nothing happens starting in vacuum. You can cook up a similar example where particles of type 2 create particles of type 1 also, and still nothing happens in vacuum, but N is not conserved (aside from N=0).
These examples are extremely easy to cook up nonrelativistically because the nonrelativistic field separates the creation and annihilation into independent fields $\psi$ and $\bar\psi$, so that $\psi$ is pure annihilation and $\bar\psi$ is pure creation, and in this Hamiltonian, I didn't put in any $\bar\psi$ terms without $\psi$, so that there is no creation starting from vacuum. If you do put it in, there are vacuum bubbles, and you get Haag's theorem again, again from the infinite volume limit.
An argument analogous to the informal Haag theorem argument works for broken symmetry too, like for nonrelativistic phonons. If you move a solid the moved solid is outside the Hilbert space, because suddenly moving the solid creates infinitely many phonons relative to the old position, because the creation amplitude is per-unit-volume.
Relativistically, all local fields include both creation and annihilation operators, so you can't introduce a local interaction without particle creation, without vacuum bubbles. If you decide to normal order the interaction, and then throw away just the terms in $:H:$ which are pure creation operators, the ones which escape the vacuum, so that you make sure the vacuum is invariant by hand, the interaction term becomes hideously nonlocal and non-Lorentz invariant, because you can't separate $\phi$ into the creation and annihilation part, $\phi^+$ and $\phi^-$ locally in a Lorentz invariant way. All vacuum bubbles contain a divergent $\delta(0)$ momentum-space delta function term, which is just the infinite volume factor due to their translation invariance.
The reason that you need to mix creation and annihilation in relativity is obvious even in pure potential scattering--- when you scatter any particle off a potential V(x) coupling to a local field, the amplitude has to be crossing invariant, because the particle propagation is not purely causal, the propagator is nonzero outside the light-cone. There is a nice popular exposition of this principle in a lecture by Feynman from 1986, which you can find online.
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