= Solution
Let $\beta=(k_BT)^{-1}$ and sum over all $2^{2L}$ configurations of the <Ising spins>. The <canonical partition function> and <Helmholtz free energy> per site are
$$
Z_L=\sum_{\{\sigma,\tau\}}e^{-\beta E(\sigma,\tau)},
\qquad \boxed{f_L=-\frac{k_BT}{2L}\log Z_L.}
$$
Both <spins> in each cell count as sites. In particular the denominator is $2L$, rather than $L$.
The mean <spin magnetization> per site is the <Boltzmann distribution> average
$$
\boxed{\langle m\rangle=\frac1{2LZ_L}\sum_{\{\sigma,\tau\}}\left[\sum_{j=1}^L(\sigma_j+\tau_j)\right]e^{-\beta E}.}
$$
Equivalently introduce a uniform <magnetic field> through $E_h=E-h\sum_j(\sigma_j+\tau_j)$ and differentiate:
$$
\langle m\rangle=\left.\frac1{2L\beta}\frac{\partial\log Z_L(h)}{\partial h}\right|_{h=0}.
$$
This convention absorbs the microscopic <magnetic moment> into $h$. Simultaneous reversal of every <spin> leaves $E$ invariant and changes the sign of the numerator, so $\langle m\rangle=0$ at zero field for finite $L$.
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