Let $M$ a manifold and $G$ a group.
Is it true that the statement: "the structure group of the frame bundle of $M$ can be reduced to $G$" simply means:
There is a subbundle $S$ of the frame bundle (i.e. at each point some distinguished set of frames, e.g. the orthogonal ones) and an atlas of $M$ such that the transition functions map $S$ to $S$?
If yes, why is this equivalent to $M/G$ having a global section?
Let $M$ be an $n$-dimensional manifold and $E_M$ it bundle of frames. Recall that if $x$ is an element of $M$, and $T_xM$ its tangent space, a linear frame at $x$ is an inveritble linear map $\mathbb{R}^n\rightarrow T_x$. We denote by $E_M(x)$ the set of linear frames at $x$, and by $E_M\cup_{x\in M}E_M(x)$, $E_M$ is a $Gl(n,\mathbb{R})$-principal bundle. It can be reduced to $G\subset Gl(n,\mathbb{R}$ is equivalent to saying that there exists a trivialisation $(U_i,g_{ij})$ of $E_M$ such that $g_{ij}:U_i\cap U_j\rightarrow G$, this implies the existence of a $G$-principal bundle $E_G\rightarrow M$ whose coordinate change are $(U_i,g_{ij})$.
Consider $E_M/G$, it is the bundle over $M$ which is obtained by gluing $U_i\times Gl(n,\mathbb{R})/G$ with $U_i\cap U_j\times Gl(n,\mathbb{R})/G \rightarrow U_i\cap U_j\times Gl(n,\mathbb{R})/G$ $(x,y)\rightarrow (x,\bar g_{ij}(x)(y))$ where $\bar g_{ij}(x)$ is the map induced by $g_{ij}(x)$ on $Gl(n,\mathbb{R})/G$ by $g_{ij}(x)$. If there exists a $G$-reduction, $g_{ij}(x)\in G$ and $\bar g_{ij}(x)$ is the identity, we deduce that $E_M/G$ is trivial, and henceforth has a global section.