= Solution
Choose the local positive $\widetilde z$ axis along the <classical spin>. For nonzero <Magnetic field>[field], a <Rotation>[proper rotation] is
$$
\boxed{S_j^x=s\widetilde S_j^z,\qquad S_j^y=\widetilde S_j^y,\qquad S_j^z=-s\widetilde S_j^x.}
$$
It preserves the <Spin commutation relations> and turns the <Magnetic field>[field] term into $-H_fS\sum_j\widetilde S_j^z$. At zero <Magnetic field>[field] choose the classical direction $+x$, so $s=1$. Expanding about the original $z$ axis without this <rotation> would put the reference state away from the classical minimum and would not give a controlled small-deviation expansion.
To the required order,
$$
\widetilde S_j^z=S-n_j,\qquad
\widetilde S_j^x=\sqrt{\frac S2}(a_j+a_j^\dagger)+O(S^{-1/2}),\qquad
\widetilde S_j^y=-i\sqrt{\frac S2}(a_j-a_j^\dagger)+O(S^{-1/2}).
$$
The longitudinal exchange gives $-JS^2+JS(n_j+n_{j+1})+O(S^0)$ on each bond. The transverse exchange gives
$$
-J\widetilde S_j^y\widetilde S_{j+1}^y
=\frac{JS}{2}\left[a_ja_{j+1}+a_j^\dagger a_{j+1}^\dagger-a_j^\dagger a_{j+1}-a_{j+1}^\dagger a_j\right]+O(S^0).
$$
Finally the <Magnetic field>[field] contributes its classical constant and $H_fSn_j$. Each site belongs to two bonds, so after removing the order-$S^2$ constant,
$$
\boxed{A=-\frac{JS}{2},\qquad B=\frac{JS}{2},\qquad C=S(2J+H_f).}
$$
These are the order-$S$ terms; the discarded products of <number operators> and the next square-root terms are order $S^0$ for <Bosonic occupation numbers>[occupations] held fixed as $S\to\infty$. A different transverse-axis phase can reverse the sign of $B$ without changing the <Energy eigenvalue>[spectrum], but the sign above follows from the stated <rotation> and oscillator convention.
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