= Solution
For the <sawtooth Ising chain>, there is a source inconsistency: the printed <Ising spin> values are $\{-1,r+1\}$, with no definition of $r$. The requested <transfer matrix> follows for $r=0$, namely the usual <Ising model> with values $\{-1,+1\}$. For arbitrary $r$, summing an apex <spin> actually gives
$$
A^{(r)}_{ss'}=e^{-\beta Jss'}\left[e^{\beta\bar J(s+s')}+e^{-\beta\bar J(r+1)(s+s')}\right].
$$
For example, $r=1$, $s=s'=-1$ and $\beta\bar J=1$ give the bracket $e^{-2}+e^4$, whereas the claimed <hyperbolic cosine> gives $e^{-2}+e^2$. Thus the printed identity is false literally. The rest of Question 1 uses the intended symmetric <Ising spin> values, $r=0$.
For a fixed configuration of base <spins>, each apex <spin> appears in only one local <Boltzmann factor>. The apex <spins> are therefore conditionally independent, and their finite sums factorize:
$$
\begin{aligned}
Z&=\sum_{s_0,\ldots,s_{L-1}=\pm1}\prod_{i=0}^{L-1}\left[e^{-\beta Js_is_{i+1}}\sum_{\bar s_i=\pm1}e^{-\beta\bar J\bar s_i(s_i+s_{i+1})}\right]\\
&=\sum_{s_0,\ldots,s_{L-1}=\pm1}\prod_i A_{s_i s_{i+1}},\qquad s_L=s_0,\\
&\boxed{A_{ss'}=2e^{-\beta Jss'}\cosh[\beta\bar J(s+s')],\qquad Z=\operatorname{Tr}A^L.}
\end{aligned}
$$
Only the base <spins> carry information from one triangle to the next. The apex <spins> have already been summed out, so no apex index remains in the <transfer matrix>; the final <trace> implements the <periodic boundary condition>.
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