In SUSY why does electroweak symmetry breaking only happen in the SM sector? This is a difficult question to phrase succinctly, so I hope the title makes sense. What I want to understand is what seems like a lack of symmetry (besides SUSY-breaking) between the SM sector and their superparters.
In SUSY we add a second higgs doublet, so we end up with 8 degrees of freedom. Three are eaten by the SM gauge bosons, leaving 5 higgs bosons. My questions is: why are the 3 degrees of freedom eaten only by SM particles rather than 3 SM and 3 SUSY particles: why the asymmetry? Would the situation be different without SUSY-breaking? 
If it helps to visualize the problem, the asymmetry is most striking if you imagine that we lived in a world where the mass scales where reversed: the SM is off at some high SUSY-breaking scale, and our world consists of superparters. Would we not have gauge theory and electroweak symmetry breaking, or would the gaugino sector require electroweak symmetry breaking? 
 A: Gauginos are spin-1/2 fermions, and they don't carry forces like the W and Z bosons do. They aren't connection coefficients, they don't superpose to macroscopic fields.
There is never a complete symmetry between bosons and fermions, even in a supersymmetric theory. The fermions are fermions and the bosons are bosons, they have completely different physical properties. The supersymmetry transformation is not like a spatial rotation--- it isn't as physical. If you rotate a sock, all the particles in the sock rotate. If you super-rotate a sock, it becomes a superposition of rotating one-particle at a time of the sock. Most of the sock stays the same, but one constituent is turned to its superpartner, and there is a quantum superposition over which constituents are flipped. The result is still mostly the original sock.
This is analogous to the notion of an infinitesimal generator, since an infinitesimal transformation acts on products one factor at a time. The SUSY transformations can be thought of as permanently infinitesimal, because their parameter squares to zero.
Supersymmetry tells you for each particle that the scattering amplitude of a boson is simply related to the scattering amplitude of the fermion. This relation is particle by particle. So supersymmetry just isn't a symmetry of objects, at least not in a useful classical sense. So in your example of the reversed heirarchy, the Higgs mechanism would still give W's and Z's a mass.
A: The quick answer to this question is: Because of the experimental data. 
Any theory beyond the SM must reproduce the results of the SM (at least to the order they have been verified by experiment). In particular, it must explain electroweak symmetry breaking (EWSB) because this is what we observe in nature. The method you describe is how EWSB is achieved when there are two higgs doublets. SUSY has two higgs doublets and therefore falls into this class of theories.
An alternative way to think about it is that EWSB happens at the scale of $\sim M_W$. Above that scale the electroweak forces are unified. SUSY however must be broken at a much higher scale (call it $M_S$) and we know this is the case because we haven't observed any super-partners in experiments so far. Since $M_S>M_W$ the electroweak forces are unified in the supersymmetric theory. We only need to break the electroweak symmetry much later after the SUSY breaking and that's why EWSB only happens in the SM model sector.
A: If you want, from a symmetry breaking point of view, the only thing that happens is that 3 goldstone bosons, coming from the broken generators are eaten, only three, and by the higgs mechanism are "eaten" by the gauge bosons, in the theory. All this in very general grounds, only taking it into account as a QFT, without worrying about the particle content. 
Also the symmetry breaking effect and the fact that the bosons are massive is going to affect the superpartners, through the radiative corrections that these "feel" from the gauge bosons. So in a way, there is a symmetry breaking everywhere in the susy sector.
