Let $(M,g)$ be a smooth Riemannian manifold. A scalar field in $(M,g)$ is merely a map $\phi:M\to \mathbb{R}$. Under a diffeomorphism $f:(M,g)\to (M,g)$ it transforms to $\phi' = \phi\circ f$. We then have the coordinate expression: $$\phi'(x')=\phi(x).\tag{1}$$
We can also say that $\phi$ is a section of the trivial bundle $M\times \mathbb{R}$ which sends $p\mapsto (p,\phi(p))$.
Now let $f:(M,g)\to (M,g)$ a conformal map with scale factor $\Omega$. This means we have $f^\ast g=\Omega^2g$. A scalar field of weight $\Delta$ transforms, in coordinates, as $$\phi'(x')=\Omega^{-\Delta}(x)\phi(x)\tag{2}.$$
My question here is: what is the underlying geometrical structure that gives rise to this transformation law? Is there some bundle in which $\phi$ is a section so that the conformal factor naturally appears when transforming the section in an appropriate way?
To compare: the transformation law of a scalar naturally follows from $\phi' = f^\ast \phi$ expressed in coordinates. More generally for tensor fields of arbitrary rank, the usual transformation law under general coordinate transformations follows from the geometric transformation using the pullback $f^\ast$ and pushforward maps $f_\ast$. I wanted to know if we have something like that for transformation under conformal maps.
What comes to mind is a possible construction of some "bundle of conformal frames", a principal ${\rm Conf}(M,g)$-bundle over $M$, such that $\phi$ is a section of some associated bundle by the appropriate representation of ${\rm Conf}(M,g)$. In particular, the dimension $\Delta$ is encoded in the representation in which $\phi$ transforms. I just don't know how to make this precise yet.
Your $\phi$ is a section of the conformal density bundle of weight $-\Delta$. Curry and Gover wrote a nice introduction to conformal geometry which explains this well (the hyperlink goes to the page where density bundles are defined).