Suppose we have a 2 site Hubbard model, with the hopping Hamiltonian given by [itex]H_t[/itex] and the Coulomb interaction Hamiltonian given by [itex]\hat{H}_U[/itex]. In the strong coupling limit (U/t >> 1), we define a canonical transformation of [itex]\hat{H} = \hat{H}_U + \hat{H}_t[/itex], as

As I understood your first post, the transformed hamiltonian in the book is explicitly written down including all second order terms and the third order terms are abbreviated O(t^3). So yes, there is a second order order term in the transformed hamiltonian but not a first order term any more as you originally claimed.

You don't want it to vanish. All I meant is, by that particular choice of operator [itex]\hat{O}[/itex] your Hamiltonian becomes [itex]H = H_U + \mathcal{O}(t^2)[/itex], which is what you deduce below.

When written in this way one can see that the [itex]\mathcal{O}(t^2)[/itex] are perturbations about the Hubbard [itex]U[/itex] term, i.e. your original system is a one with all electrons frozen at their lattice sites and then your introduce hopping by the [itex]t[/itex] terms. In fact the [itex]\mathcal{O}(t^2)[/itex] term is the Heisenberg model and represents super exchange which leads to effective spin flips.