The proof given in Iwaniec-Kowalski is, as it stands, wrong. It can be easily fixed, as I explain below.
In general, one can think of $\nu(n)^2$ as the characteristic function of integers with $P^-(n):=\min(p|n)\ge y$. So
$$
\sum_{n\le x} \frac{\nu(n)^2 f(n)}{n} \approx \sum_{ n\le x,\ P^-(n)\ge y } \frac{f(n)}{n} \asymp \prod_{y\le p\le x} \left(1+\frac{f(p)}{p}\right)
$$
for any reasonable multiplicative function $f$ that is bounded on primes. However, an important restriction is that $f(p)<\kappa$ on average, where $\kappa$ is the sifting dimension. In this case the dimension is 1, whereas $f(p)=8$. So the sum in question, say $S$, does not satisfy a priori the claimed bound. In fact, in this case $S\gg(\log x/\log y)^8$:
If $P^+(n)=\max(p|n)$, then we have that
$$
S= \sum_{n\le x}\frac{\nu(n)^2\tau(n)^3}{n}
\asymp \sum_{P^+(n)\le x}\frac{\nu(n)^2\tau(n)^3}{n}
$$
(this step is heuristic and employed for simplicity). If we let $\sigma_m=\sum_{[d_1,d_2]=m}\theta_{d_1}\theta_{d_2}$ and we open $\nu(m)^2$, we have that
$$
S \approx \sum_{P^+(m)\le x} \sigma_m \sum_{P^+(n)\le x,\ m|n} \frac{\tau(n)^3}{n}
= P(x) \sum_{P^+(m)\le x} \frac{\sigma_m g(m)}{m},
$$
where $g(m)$ is a multiplicative function with $g(p)=8+O(1/p)$ and $P(x)=\prod_{p\le x}(1+\tau(p)^3/p+\tau(p^2)^3/p^2+\cdots)\asymp(\log x)^8$. Hence
$$
S/P(x)\approx \sum_{P^+(d_1d_2)\le x} \frac{\theta_{d_1}\theta_{d_2}g([d_1,d_2])}{[d_1,d_2]}
= \sum_{P^+(d_1d_2)\le x} \frac{\theta_{d_1}\theta_{d_2}g(d_1)g(d_2)}{d_1d_2} \frac{(d_1,d_2)}{g((d_1,d_2))}.
$$
Writing $f(m)=\prod_{p|m}(1-g(p)/p)$ so that $m/g(m)=\sum_{n|m}f(n)n/g(n)$, we deduce that
$$
S/P(x)\approx \sum_{P^+(n)\le x}\frac{f(n)n}{g(n)} \left( \sum_{P^+(d)\le x,\ n|d} \frac{\theta_{d}g(d)}{d}\right)^2.
$$
When $y/2 \lt n\le y$, we have that
$$
\sum_{P^+(d)\le x,\ n|d} \frac{\theta_{d}g(d)}{d}
= \frac{\theta_n g(n)}{n} = \frac{\mu(n)g(n)}{G} \sum_{k\le y,\ (k,q)=1,\ n|k} \frac{\mu^2(k)}{\phi(k)}
= \frac{\mu(n)g(n)}{G} \cdot \frac{\textbf{1}_{(n,q)=1}}{\phi(n)}
$$
(note that there is an error in the definition of $\theta_b$ in Iwaniec-Kowalski, as one has to restrict them on those $b$ which are coprime to $q$. In fact, $\theta_b=(\mu(b)b/G) \sum_{k\le y,\ (k,q)=1,\ b|k}\mu^2(k)/\phi(k)$). So
$$
S \gtrsim \frac{P(x)}{G^2} \sum_{y/2 \lt n\le y,\ (n,q)=1} \frac{\mu^2(n) f(n)g(n)n}{\phi(n)^2}
\gg_q (\log x)^8(\log y)^5,
$$
by the Selberg-Delange method. This is certainly bigger than what we would need.
In order to remedy this deficiency, one would have to choose $\nu(n)$ in another way, as weights from a sieve of dimension 8 or higher. The easiest choice to work with is, as GH also points out, is to define $\nu$ via the relation $\nu(n)^2=(1*\lambda)(n)$, where $(\lambda(d):d\le D)$ is a $\beta$ upper bound sieve of level $y$ and dimension 8, so that
$$
S = \sum_{n\le x} \frac{(1*\lambda)(n)\tau(n)^3}{n}.
$$
The point is that the sequence $(\lambda(d))_{d\le D}$ satisfies the Fundamental Lemma (Lemma 6.3 in Iwaniec-Kowalski) in the following strong sense:
(1) $\lambda(d)$ is supported on square-free integers with $P^+(d)\le y$
(2) whenever $\{a(d)\}_{d\ge1}$ is a sequence such that
$$
\bigg|\sum_{P^+(d)\le z}\frac{\mu(d)a(dm)}{d}\bigg|
\le Ag(m)\prod_{y_0\le p\le z}(1-g(p)/p)
\quad\text{when}\ P^-(m)>z,
$$
where $A\ge0$ and $y_0\ge1$ are some parameters and $g\ge0$ is multiplicative with
$$
\prod_{w\le p\le w'} (1-g(p)/p)^{-1} \le \left(\frac{\log w'}{\log w}\right)^8\left(1+\frac{K}{\log w}\right)\qquad(w'\ge w\ge y_0),
$$
we have
$$
\sum_{d\le D} \frac{\lambda(d)a(d)}{d} = \sum_{P^+(d)\le y} \frac{\mu(d)a(d)}{d}
+ O_K\left( Ae^{-u}\prod_{y_0\le p\le y}\left(1-\frac{g(p)}{p}\right) \right),
$$
where $u=\log D/\log y$.
Remark: if $a=g$, then the condition on $a$ holds trivially with $A=y_0=1$. Hence, the above statement is indeed a generalization of the usual Fundamental Lemma.
We have that
$$
S = \sum_{d\le D} \frac{\lambda(d)a(d)}{d},
$$
where
$$
a(d) = \sum_{m\le x/d} \frac{\tau(dm)^3}{m}.
$$
If $P^-(m)>z$, then
\begin{align*}
\sum_{P^+(d)\le z} \frac{\mu(d)a(dm)}{d}
&= \sum_{P^+(d)\le z} \frac{\mu(d)}{d}\sum_{k\le x/(dm)}\frac{\tau(dkm)^3}{k} \\
&= \sum_{n\le x/m} \frac{\tau(nm)^3}{n} \sum_{dk=n,\,P^+(d)\le z}\mu(d).
\end{align*}
By M\"obius inversion, we then conclude that
\begin{align*}
\sum_{P^+(d)\le z} \frac{\mu(d)\tau(d)^3a(dm)}{d}
&= \sum_{n\le x/m,\,P^-(n)>z} \frac{\tau(nm)^3}{n} \\
&\ll \tau(m)^3 (\log x/\log z)^8 \\
&\asymp (\log x)^8\tau(m)^3\prod_{11\le p\le z}(1-\tau(p)^3/p),
\end{align*}
so that the second axiom holds for $a$ with $A\asymp(\log x)^8$, $y_0=11$ and $g=\tau^3$. We thus conclude that
\begin{align*}
S = \sum_{d\le D} \frac{\lambda(d)a(d)}{d}
&= \sum_{P^+(d)\le y} \frac{\mu(d)a(d)}{d}
+ O\left( e^{-\log x/\log y} \left(\frac{\log x}{\log y}\right)^8\right) \newline
&= \sum_{n\le x,\ P^-(n)>y} \frac{\tau(n)^3 }{n}
+ O\left( e^{-\log x/\log y} \left(\frac{\log x}{\log y}\right)^8\right) \newline
&\ll \left(\frac{\log x}{\log y}\right)^8.
\end{align*}
A more general remark: Selberg's sieve is not as flexible as the $\beta$-sieve as far as ``preliminary sieving'' is concerned because it carries inside it the sieve problem it is applied to, in contrast to the $\beta$-sieve weights that only depend on the sifting dimension via the $\beta$ parameter.
EDIT: the monotonicity principle II described in page 49 of "Opera de Cribro" by Friedlander-Iwaniec is a way to use Selberg's sieve weights for preliminary sieving. See also Proposition 7.2 in the same book.
Best Answer
In general, the partition function of the Ising model is usually nicer when defining $x = \tanh(\beta)$ instead of $t = \exp(- \beta)$. One explanation for this is that the Ising model partition function is the same as the partition function for the random cluster model and the partition function of the Ising model can be calculated using the loop-O(1)-model with parameter $x = \tanh(\beta)$ and Bernoulli percolation with parameter $x$.
I tried to adopt this point of view here: https://alea.impa.br/articles/v19/19-07.pdf
Let me give just an example with the graph which is just a loop of $n$ vertices and $n$ edges. Then I can find the partition function by taking all even (spanning) subgraphs of the graph (which means subsets of edges such that all vertices have even degree.) In this particular case it is the full subgraph and the empty one. Then the partition function of the Ising model is given as
$$ Z = \sum_{g \text{ even subgraph of } G} x^{o(g)} $$ where $o(g)$ is the number of open edges of $g$ meaning that $o(\emptyset) = 0$ and $o(G) = n$ in the particular case.
Thus, we have that $ Z = 1 + x^n$ in that particular case. I do not really understand the definition of your graphs, but this point of view might be helpful.