In the answer of K. Conrad to this question, he mentions a "nice 4-term short exact sequence of abelian groups (involving units groups mod a, mod b, and mod ab)" proving the product formula for $\phi$. How does this sequence look like? (I couldn't figure it out myself, the term $(a,b)$ in the formula puzzles me.) In a sense he only mentions the obvious about the sequence, that $U(a)$, $U(b)$ and $U(ab)$ should occur (which is not surprising since by definition $\phi(a):=\#U(a)$).
EDIT: Perhaps the sequence might be $$0\rightarrow \ker\rightarrow U(\mathbf{Z}/ab\mathbf{Z})\rightarrow U(\mathbf{Z}/a\mathbf{Z})\times U(\mathbf{Z}/b\mathbf{Z})\rightarrow \operatorname{Coker}\rightarrow 0$$
As KCd mentions in the comments on his answer, the kernel and cokernel should be made more explicit. In your follow-up question I explicitly write down the elements in the kernel, and mention how the cokernel should be written as $U(\gcd(a,b))$, where the map is $(a,b)\mapsto ab^{-1}$. So we get
$$1\to K\to U(ab)\to U(a)\times U(b)\to U(\gcd(a,b))\to 1$$
as a short exact sequence (thanks to CRT). By the alternating product formula, we obtain
$$\frac{\gcd(a,b)\varphi(a)\varphi(b)}{\varphi(ab)\varphi(\gcd(a,b))}=1.$$