= Solution
\b[The sum needs a stated interpretation.] The original equation leaves the second bond outside explicit parentheses, with its index then formally free. The natural <Translation invariance>[translation-invariant], equal-bond interpretation used here is the <energy> $-J\sum_j\tau_j(\sigma_j+\sigma_{j+1})$. The subsequently mentioned $h$ has no term in that <energy>, so no <magnetic field> or <single-ion anisotropy> is added. Such an extra term would define another model and change the answer.
Let $K=\beta J$. Sum over each decorating <spin> before summing over the remaining <spins>. The <decimation of an alternating mixed-spin chain> gives the effective <transfer matrix>:
$$
V_{st}=\sum_{t_0=-1}^{1}e^{Kt_0(s+t)}=1+2\cosh[K(s+t)]
=\begin{pmatrix}1+2\cosh2K&3\\3&1+2\cosh2K\end{pmatrix}_{st}.
$$
No on-site entropy has been omitted: the term $1$ is the weight of the zero decorating <spin>. The <Normalization>[normalized] <eigenvectors> are $v_\pm=(1,\pm1)^{\mathsf T}/\sqrt2$, with <eigenvalues>
$$
\boxed{\lambda_+=4+2\cosh2K=2+4\cosh^2K,\qquad \lambda_-=2\cosh2K-2=4\sinh^2K.}
$$
There are $L$ transfer steps but $2L$ physical sites, so
$$
\boxed{Z_L=\lambda_+^L+\lambda_-^L,\qquad
f=\lim_{L\to\infty}\frac{-k_BT\log Z_L}{2L}=-\frac{k_BT}{2}\log(4+2\cosh2\beta J).}
$$
The <free energy> per two-site <unit cell> is twice this value. At $J=0$, the formula gives $Z_L=6^L$, as required by two states on one sublattice and three on the other. As $T\to0$, the <free energy> per physical site tends to $-J$, the <energy> of an aligned chain with two satisfied bonds per <unit cell>. These checks also fix the factor of two in the site count.
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