= Solution
The second-order terms are
$$
[S,H_t]+\frac12[S,[S,\mathcal E]]
=[S,H_t]-\frac12[S,H_t],
$$
so
$$
\boxed{H'=\mathcal E(\Delta)+H^{(2)}+O(t^3),\qquad
H^{(2)}=\frac12[S,H_t].}
$$
Here $S$ is the first-order generator. The full second-order <commutator> includes terms changing the number difference by four. To retain only fixed site occupancies, keep the products $XY$ and $YX$.
With $f=f(\Delta)$ and $f_+=f(\Delta+2)$, the mixed <commutators> give
$$
[fX,Y]=fXY-f_+YX,\qquad
[-f_+Y,X]=fXY-f_+YX.
$$
Thus <occupancy-preserving second-order bosonic hopping> gives
$$
\boxed{H^{(2)}_{\rm occ}
=t\left[f(\Delta)XY-f(\Delta+2)YX\right].}
$$
For <bosons>, <normal ordering> gives
$$
XY=N_a+P,\qquad YX=N_b+P,
\qquad P=\sum_{s,s'}a_s^\dagger a_{s'}b_{s'}^\dagger b_s.
$$
Consequently an explicit form in terms of the supplied function is
$$
\boxed{H^{(2)}_{\rm occ}
=t\left[f(\Delta)N_a-f(\Delta+2)N_b
+\{f(\Delta)-f(\Delta+2)\}P\right].}
$$
The operator $P$ exchanges <spin> labels between sites while preserving their total occupancies. Defining dimensionless Schwinger-boson <spins> $\mathbf s_a=\tfrac12a_s^\dagger\boldsymbol\sigma_{ss'}a_{s'}$ and similarly for $B$,
$$
P=\frac12N_aN_b+2\mathbf s_a\cdot\mathbf s_b.
$$
No factor of $\hbar^2$ is needed for these dimensionless <spins>; for physical <spin> operators $\mathbf S_a=\hbar\mathbf s_a$ it becomes $2\mathbf S_a\cdot\mathbf S_b/\hbar^2$. The <spin> exchange belongs in the occupancy-preserving part, whereas two hops in the same direction do not.
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