Proving formula involving Euler's totient function

I'll adjust your notation a bit, using $x\in U_{mn}$ for an invertible element of $\mathbb{Z}/mn\mathbb{Z}$, using $\bar x\in U_m$ for the residue class of $x$ modulo $m$, and using $\tilde x \in U_n$ for the residue class of $x$ modulo $n$.

The image of your map $x \mapsto (\bar x,\tilde x)$ is generally smaller than $U_m \times U_n$ because $\bar x$ and $\tilde x$ will always be the same modulo $d$. We first choose a reduced residue system $\{a_1=1, a_2, \ldots, a_{\varphi(d)}\}$ modulo $d$ from the elements $U_{n}$ and consider the images of the maps $f_i: x \mapsto (\bar x, a_i\tilde x)$. (Note that each $a_i$ is invertible modulo $n$.) It's clear that the images of these maps are disjoint, have the same size, and that we are studying the special case $f_1:x \mapsto (\bar x, \tilde x)$. In fact, the union of these images is all of $U_m \times U_n$, as we now show.

Take any $(y,z) \in U_m \times U_n$. We show it is equal to some $f_i(x)$ where $a_i \equiv zy^{-1} \pmod{d}$. By a slight generalization of the Chinese Remainder Theorem, there is a unique $x$ modulo $\frac{mn}{d}$ such that $$x \equiv y \pmod{m}\qquad \qquad \text{and} \qquad \qquad x \equiv z{a_i}^{-1} \pmod{n}.$$ Then $f_i(x)=(\bar x, a_i \tilde x) =(y,z)$. (In fact, the $d$ preimages of $(y,z)$ are the elements $x+\frac{mn}{d} k$ with $0 \leq k \leq d-1$.)

A generalization of the Chinese Remainder Theorem is required because $m$ and $n$ share the factor $d$. Such systems of congruences have a solution as long as they are compatible modulo $d$, and this solution is unique modulo $\rm{lcm}(m,n)=\frac{mn}{d}$. Our system is compatible modulo $d$, since $y \equiv z {a_i}^{-1} \pmod{d}$.

Thus, the index of the image of $x \mapsto (\bar x, \tilde x)$ is $\varphi(d)$.

Jonas Kibelbek directly proved that the index of the image of $\eta$ is $\phi(d),$ and below is an alternative by proving an exact sequence, which I hope might clarify the matter somewhat.

The exact sequence I want to prove is

$$0\rightarrow \text{Ker}(f)\rightarrow U_{mn}\overset{f}{\rightarrow}U_m\times U_n\overset{g}{\rightarrow} U_d\rightarrow0,$$

where $f(x+mn\mathbb Z)=(x+m\mathbb Z, x+n\mathbb Z),$ and $g(x+m\mathbb Z, y+n\mathbb Z)=xy^{-1}+d\mathbb Z$ (Here the inverse is taken modulo $d$).
Proof:
Firstly, $\forall (a+d\mathbb Z)\in U_d,$ we have that $g(a+m\mathbb Z, 1+n\mathbb Z)=(a+d\mathbb Z),$ so $g$ is surjective. Further, it is clear that $g\circ f$ vanishes. Conversely, if $(x+m\mathbb Z, y+n\mathbb Z)\in U_m\times U_n$ is such that $x\equiv y\pmod d,$ then, by a slight generalisation of Chinese rmainder theorem, as in Kibelbek's answer, $\exists z$ such that $\begin{cases}z\equiv x\pmod m\\z\equiv y\pmod n\end{cases}.$ Thus the sequence is exact. Q.E.D.

P.S. This sequence is in essence the sequence in this answer, with some reductions and modifications; I mark this answer as CW, for there is nothing new in this answer.