Let $D \subseteq \mathbb{C}$, $\omega = P dx+Q dy$ $1-$form over $D$, is said "of class $C^{1}$", if $P,Q : D \longmapsto \mathbb{C}$ are of class $C^{1}$.
In this case we define formally $d\omega = (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dxdy$
I do understand that with this definition it follows almost directly the following particular case of Green Stokes theorem as a consequence of Fubini Tonelli :
$\textbf{Theorem :}$ $\omega$ $1-$form of class $C^{1}$ over a right-angled $R \subseteq D$, then $\int_{\partial R} \omega = \int_{R} d\omega$
But are there any other reason to this definition ? In my eyes could just be seen a definition to prove the theorem, what I would like, is to find a deeper meaning which makes the definition well defined, and useful.
You do not "formally define" $d\omega$ in that way. Let $\omega = P dx + Qdy$ as you have. This is a $1$--form, the space of $1$--forms is commonly denoted $\Omega^1$. The exterior derivative $d : \Omega^1 \to \Omega^2$ takes is defined by $$d\omega = \left( \frac{\partial P}{dx} dx \right) \wedge dx + \left( \frac{\partial Q}{dx} dx \right) \wedge dy + \left( \frac{\partial P}{dy} dy \right) \wedge dx + \left( \frac{\partial Q}{dy} dy \right) \wedge dy.$$ Since $dx \wedge dx =0$, and $dy \wedge dy =0$ , we see that the first and last terms are zero. Moreover, since $dx \wedge dy = - dy \wedge dx$, we get $$d \omega = \left( \frac{\partial Q}{dx} - \frac{\partial P}{dy} \right) dy \wedge dx.$$
A clean exposition of the proof of Green's theorem is given by Joyce: https://mathcs.clarku.edu/~djoyce/ma131/greenproof.pdf