I'm interested in the OP's curiosity. I'd agree with Waylander about disfavoring to make 4-membered rings. Yet, it is possible to have two six-membered ring products. The first $\alpha$-proton abstraction followed by Michael addition resulted the first intermediate as OP suggested:
Now, this newly formed intermediate contain 4 types of eligible $\alpha$-protons, namely, $\ce{H_a, H_b, H_c,}$ and $\ce{H_d}$. Amongst them, only $\ce{H_a}$ is primary, while $\ce{H_c}$ is tertiary and the rest, $\ce{H_b}$ and $\ce{H_d}$, are secondary. The base would abstract each proton in rates according to the dissociation energy of each (e.g., the fastest rate with tertiary and the slowest with primary) but all abstractions are in fast equilibrium as shown in the scheme below:
$\ce{H_a}$ and $\ce{H_d}$ proton abstractions resulted the formation of intermediates $\bf{I}$ and $\bf{II}$, which would abstract from a proton from water to give the corresponding aldols. Intermediate $\bf{I}$ is the most stable among others and resulted aldol would undergo further dehydration to give the corresponding $\alpha,\beta$-unsaturated ketone as the major product (as correctly suggested by the OP). Meanwhile, formation of the intermediate $\bf{II}$ is slow compared to that of $\bf{I}$ due to greater ring strain (even though it is a six-membered ring formation, the resulting intermediate is a bicyclic product with a bridged-head). Nevertheless, some would form to give the corresponding aldol as a minor product, which would not undergo dehydration because $\alpha$-hydrogen is locked in a position that restrict the elimination.
On the other hand, although abstractions of $\ce{H_b}$ and $\ce{H_c}$ protons are possible and in fast equilibrium, the formation of intermediates $\bf{III}$ and $\bf{IV}$ are very minute depending on the corresponding activation energies. The rates of these formations are very slow or not exist compared to other two, mostly due to very unfavorable 4-membered ring formation.
