Elementary Proof
Let $p$ and $q$ be coprime. The Chinese remainder theorem defines a bijection $\psi:\mathbb{Z}_p\times\mathbb{Z}_q\rightarrow\mathbb{Z}_{pq}$ given by $\psi(a,b)=qa+pb\pmod{pq}$. The crux, and missing detail of the proof in your book is that (LEMMA) for any $(a,b)\in\mathbb{Z}_p\times\mathbb{Z}_q$, we have that $\gcd(qa+pb,pq)=1$ iff $\gcd(a,p)=\gcd(b,q)=1$. This lemma eqivalently tells us that $\psi$ restricts to a bijection between from the set $\{(a,b)\in\mathbb{Z}_a\times\mathbb{Z}_b:\gcd(a,q)=\gcd(b,p)=1\}$ to the set $\{c\in\mathbb{Z}_{pq}:\gcd(c,pq)=1\}$, and thus that $\varphi(p)\varphi(q)=\varphi(pq)$. We prove this lemma below.
First, we prove $(\Rightarrow)$. Let $\gcd(qa+pb,pq)=1$, then if $n$ is a divisor of both $a$ and $p$, it must also divide $qa+pb$ and $pq$, so $n$ would have to divide $\gcd(qa+pb,pq)=1$, meaning that $n=1$. Since the only natural $n$ that divides both $a$ and $p$ is $n=1$, then $\gcd(a,p)=1$. We can repeat this same argument to show that $\gcd(b,q)=1$.
We now prove $(\Leftarrow)$. Let $\gcd(a,p)=\gcd(b,q)=1$. Since $q,a$ are both coprime to $p$ and $p,b$ are both coprime to $q$, then $\gcd(qa,p)=\gcd(pb,q)=1$. Since $p,q$ are coprime, then
\begin{equation}
\begin{split}
\gcd(qa+pb,pq)&=\gcd(qa+pb,p)\gcd(qa+pb,q)\\
&=\gcd(qa,p)\gcd(pb,q)\\
&=1
\end{split}
\end{equation}
by the properties of the $\gcd$. This completes our proof of the lemma.
More Abstract Proof
Let's first define a useful concept. For any ring $R$, we may define $R^\times$ to be the set of units (elements with an inverse) of $R$; this is in fact a group under the multiplication operator inherited from $R$, and is known as the unit group of $R$.
It is not hard to see that the units of $\mathbb{Z}_n$ are simply the elements $[k]\in\mathbb{Z}_n$ such that $k$ is coprime to $n$. In other words, $|\mathbb{Z}_n^\times|=\varphi(n)$.
Now, the Chinese remainder theorem tells us that if $p,q$ are coprime, then there exists a ring isomorphism $\psi:\mathbb{Z}_{pq}\rightarrow\mathbb{Z}_p\times\mathbb{Z}_q$. Since $\psi$ is a ring isomorphism, then it must also restrict to a group isomorphism $\mathbb{Z}_{pq}^\times\rightarrow(\mathbb{Z}_p\times\mathbb{Z}_q)^\times$. Combining this with the fact that $(\mathbb{Z}_p\times\mathbb{Z}_q)^\times\cong\mathbb{Z}_p^\times\times\mathbb{Z}_q^\times$, we have that $\mathbb{Z}_{pq}^\times\cong\mathbb{Z}_p^\times\times\mathbb{Z}_q^\times$, and thus
\begin{equation}
\varphi(pq)=|\mathbb{Z}_{pq}^\times|=|\mathbb{Z}_p^\times|\cdot|\mathbb{Z}_q^\times|=\varphi(p)\varphi(q)
\end{equation}
as desired.
