First I have the definition of a conformal map. Let $(M,g)$ and $(M',g')$ be two pseudo-Riemannian manifolds of same dimension. Let $U\subset M$ and $V\subset M'$, we say that a smooth map of maximal rank $\Phi : U\to V$ is a conformal map if there is some smooth $\Omega : U\to [0,+\infty)$ such that $$\Phi^\ast g'=\Omega^2 g.\tag{1}$$
Now, intuitively I would imagine that by a global conformal transformation one means just a globally defined conformal map $\Phi : M\to M'$.
My intuition would tell that to define the global conformal group of $M$ we would just need to take all globally defined conformal diffeomorphisms $\Phi : M\to M$.
Still one usually introduces a conformal compactification here. Take for instance this Phys.SE post or rather the book "A Mathematical Introduction to Conformal Field Theory" by M. Schottenloher. A canonical reference taking the same approach is the paper "Relativistic Symmetry Groups" by Penrose.
All these authors are talking about $(M,g) = \mathbb{R}^{p,q}$. To define the global conformal group they pick $\overline{\mathbb{R}^{p,q}}$ the conformal compactification and define $\operatorname{Conf}(p,q)$ the group of globally defined conformal transformations on it.
Why is that? Why not just define $\operatorname{Conf}(p,q)$ to be the group of globally defined conformal diffeomorphisms on $\mathbb{R}^{p,q}$ itself, i.e., maps $\Phi : \mathbb{R}^{p,q}\to \mathbb{R}^{p,q}$ satisfying Eq. (1) with the canonical metric tensor? Why pass to the conformal compactification?