For each real vector bundle of rank $n$, $\zeta: p: V \to X$. An orientation is a continuous choice of orientations of each fiber. Let $\mathbb{R}^{n}$ is given the standard orientation and denote $Or(\mathbb{R}^{n})$ is two-point set which contains two orientation of $\mathbb{R}^{n}$ then we may define the two-fold covering space:
$$Or(\zeta):V_{GL} \times_{GL(n,\mathbb{R})} Or(\mathbb{R^{n}}) \to X$$
where $V_{GL}$ is associated principal $GL(n,\mathbb{R})$- bundle and $GL(n,\mathbb{R})$ acts on $Or(\mathbb{R}^{n})$ by matrix multiplication. I want to ask how an orientation of $\zeta$ is equivalent to a choice of section of the bundle $Or(\zeta)$.
Firstly, remark that $Gl(\mathbb{R}^n)\times_{Gl(n,\mathbb{R}^n}Or(\mathbb{R}^n)$ has two elements. Let $a,b$ denote the two elements of $Or(\mathbb{R}^n)$, and $U\in Gl(\mathbb{R}^n)$, $(U,a)=U(I,a)$ if $U$ preserves the orientation, or $(U,a)=(I,b)$ if $U$ does not preserve the orientation. You also have $(U,b)=U(I,b)$ or $(U,b)=U(I,a)$. This implies that the two elements of the quotient space $Gl(\mathbb{R}^n)\times_{Gl(n,\mathbb{R}^n}Or(\mathbb{R}^n)$ are the classes of $(I,a)$ or $(I,b)$.
This implies that $Or(\zeta)$ is a bundle whose fibre has $2$-elements. (I assume that $M$ is connected). It is also a principal bundle in fact its fibre is $\mathbb{Z}/2$ (The multiplication of matrices induces the group structure on the fibre). You deduce that if it has a section, it is trivial, and the section is a map $s:M\times M\times Or\mathbb{R}^n)$ defined by $s(x)=(x,u(x))$, $u$ is constant since $M$ is connected and the orientation is well-defined.
Conversely, an orientation of $\zeta$ is a map $u:M\rightarrow Or(V_x)$ which defines a section $x\rightarrow (x,u(x))$ of $Or(\zeta)$.