Basic question:
If we have $$Y=d\left(\frac{1}{\alpha} +\frac{1}{\bar{\alpha}}\right)$$ where $d$ is exterior derivative, i.e, $Y$ is a $1$-form.
Now we could write that as
$$Y=d\left(\frac{\bar{\alpha}}{|\alpha|^2} + \frac{\alpha}{|\alpha|^2}\right)$$
where $|\alpha|^2=\alpha \bar{\alpha}$.
Can we simply the last equation to
$$Y=\frac{1}{|\alpha|^2}d(\bar{\alpha}+\alpha)$$ or the presence of the exterior derivative does not allow us to move the now real function, $|\alpha|^2$, outside it?
No, you cannot do that as the exterior derivative is not linear over functions. In general, if $f$ is a smooth function and $\omega$ is a $k$-form, $d(f\omega) = df\wedge\omega + fd\omega$ which you'll note is not the same thing as $fd\omega$.