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). Opening $\nu(m)^2$ using GH's notation, 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 $P(x)=\prod_{p\le x}(1+\tau(p)^3/p+\tau(p^2)^3/p^2+\cdots)\asymp(\log x)^8$. Writing $\sigma_m=\sum_{[d_1,d_2]=m}\theta_{d_1}\theta_{d_2}$, we find that
$$
S/P(x)\approx \sum_{P^+(d_1d_2)\le x} \frac{\theta_{d_1}\theta_{d_2}g([d_1,d_2])}{[d_1,d_2]}
= \sum_{P^+(d_1d_2)\le x} \frac{\theta_{d_1}\theta_{d_2}g(d_1)g(d_2)}{d_1d_2} \frac{(d_1,d_2)}{g((d_1,d_2))}.
$$
Writing $f(m)=\prod_{p|m}(p/g(p)-1)$ so that $m/g(m)=\sum_{n|m}f(n)$, we deduce that
$$
S/P(x)\approx \sum_{P^+(n)\le x}f(n) \left( \sum_{P^+(d)\le x,\ n|d} \frac{\theta_{d}g(d)}{d}\right)^2.
$$
When $y/2 \lt n\le y$, we have that
$$
\sum_{P^+(d)\le x,\ n|d} \frac{\theta_{d}g(d)}{d}
= \frac{\theta_n g(n)}{n} = \frac{\mu(n)g(n)}{G} \sum_{k\le y,\ (k,q)=1,\ n|k} \frac{\mu^2(k)}{\phi(k)}
= \frac{\mu(n)g(n)}{G} \textbf{1}_{(n,q)=1}
$$
(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
$$
S \gtrsim \frac{P(x)}{G^2} \sum_{y/2 \lt n\le y,\ (n,q)=1} \frac{\mu^2(n) h(n)g(n)^2}{\phi(n)^2}
\gg_q (\log x)^8(\log y)^5,
$$
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) if $|a(d)|\le A$ for all $d\le D$ and $g(d)$ is multiplicative with
$$
\prod_{w\le p\le z} (1-g(p)/p)^{-1} \le \left(\frac{\log z}{\log w}\right)^8\left(1+\frac{K}{\log w}\right),
$$
then
$$
\sum_{d\le D} \frac{\lambda(d)a(d)g(d)}{d} = \sum_{P^+(d)\le y} \frac{\mu(d) a(d) g(d)}{d}
+ O_K\left( Ae^{-u}\prod_{p\le y}\left(1-\frac{g(p)}{p}\right) \right),
$$
where $u=\log D/\log y$.

We have that
$$
S = \sum_{d\le D} \frac{\lambda(d)\tau(d)^3 a(d)}{d},
$$
where
$$
a(d) = \sum_{m\le x/d} \frac{\tau(dm)^3}{\tau(d)^3m} \ll(\log x)^8
$$
for $d\le x$. So

\begin{align*}
S = \sum_{d\le D} \frac{\lambda(d)\tau(d)^3 a(d)}{d}
&= \sum_{P^+(d)\le y} \frac{\mu(d)\tau(d)^3 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*}

where the second equality follows by M\"obius inversion.

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.