It sounds like you're misunderstanding the proof: the main theorem is

There is a computable function $f$ (such things are in general called *many-one reductions*) from $\mathbb{N}$ to $\{$finite sets of tiles$\}$ (construed appropriately) such that $\Phi_e(e)\uparrow$ if and only if the set of tiles $f(e)$ can tile the plane.

From this we conclude that the set of finite sets of shapes which tile the plane is not computable, since otherwise - pulling back along $f$ - we could compute the halting problem.

At this point, maybe (I haven't read the proof in detail in a long time) it turns out that every set of tiles of the form $f(e)$ which *does* tile the plane, does so in a "hierarchical" fashion (whatever that means). This might be the case; however, it would be wrong to say that the proof "only works for hierarchical tilings," since the tilings aren't the 'inputs' to the theorem. Additionally, I'd be very surprised if this were true: I have a vague picture in my head of what "hierarchical" ought to mean to me, and my recollection is that the tilings constructed in the proof of undecidability of the tiling problem appear to be *not* hierarchical.

It might be the case that all the images of all such reductions $f$ all have to mostly consist of hierarchically tilable sets of tiles; for example, it is conceivable that

There is no computable function $f$ from $\mathbb{N}$ to $\{$finite sets of tiles$\}$, such that (i) $f(e)$ can tile the plane if and only if $\Phi_e(e)\uparrow$, and (ii) each set of tiles $f(e)$ (for $\Phi_e(e)\uparrow$) can *not* tile the plane hierarchically.

Depending on the definition of "hierarchical," I'd be surprised by that result, but in principle it could be true. However, if true it would require considerable work to prove, probably more than proving the original result.

None of the above, by the way, relies on aperiodicity. Where aperiodicity arises is in the fact that, *if* every set of tiles which could tile the plane could do so periodically, *then* the tiling problem would be computable (see the bottom of the first page of the article), so we must be using mostly sets of tiles which do not tile the plane periodically. That is, a lemma standing 'morally' (at least I think so) at odds to the undecidability of the tiling problem is:

There is no computable function $f$ from $\mathbb{N}$ to $\{$finite sets of tiles$\}$ such that $(i)$ $f(e)$ can tile the plane if and only if $\Phi_e(e)\uparrow$, and $(ii)$ if $\Phi_e(e)\uparrow$ then $f(e)$ can tile the plane periodically.

So aperiodicity is really an after-the-fact thing: *because* the tiling problem is unsolvable, it *therefore follows* that there are finite sets of tiles which can tile the plane, but can not do so periodically. (This is the first full sentence on page 178 of Robinson's article.)

I'm a little unclear, though, exactly what your question is, so I might not have addressed it well; is this what you were looking for?