I think the order structure you describe is identical to the face lattice of the $d$-cube. You then ask about an order preserving and surjective map between two such lattices, but not in a standard way, that is, e.g. not like an order embedding. It looks more like you try to find the face lattice of $Q_d$ as a "sub-lattice" in $Q_{d+k}$, but in kind of a graph minor sense.
The set $S(Q_d)$ can also be identified with $\{*,0,1\}^d$, i.e., the set of sequences of length $d$ with these three symbols. One way to map $S(Q_{d+k})\to S(Q_d)$ surjectively and order-preservingly is by restriction, that is, e.g. by deleting the $k$ first entries of the sequence. This corresponds to the geometric operation of projection, as we project the $(d+k)$-cube onto one of its $d$-dimensional faces (which is a $d$-cube).
There are other such maps, though. Look at the following picture, which is meant to partially visualize a map $S(Q_3)\to S(Q_2)$.
This map can be completed to a projection onto some face of $Q_3$ (the face is not uniquely determined). But there are other ways to complete this map, which does not correspond to such a projection:
- The red vertices of $Q_3$ are mapped to the red vertices of $Q_2$.
- The blue edges of $Q_3$ are mapped to the blue edges of $Q_2$.
- The green faces of $Q_3$, their edges, and their non-red vertices are mapped to the green edges of $Q_2$.
- All other subcubes of $Q_3$ ($Q_3$ itself, the non-green faces and the the non-blue edges) are mapped to $Q_2$ itself.
This is not a projection onto a face, since in such a projection from $Q_3$ to $Q_2$ only two faced would be mapped to all of $Q_2$. But you can view this as a diagonal projection as visualized in the following picture.
Similar diagonal projections are certainly possible when going from $Q_{d+1}$ to $Q_d$. And composing serveral such maps then gives you non-standard projections from $Q_{d+k}$ to $Q_d$.