[step:Transport the standard orientation through an atlas with positive transition determinants]Assume first that there is a bundle atlas $\{(V_a,\Psi_a)\}_{a\in A}$ with transition maps $h_{ab}:V_a\cap V_b\to GL(k,\mathbb R)$ satisfying $\det h_{ab}(x)>0$ for every $x\in V_a\cap V_b$.
For each $a\in A$ and each $x\in V_a$, define an ordered basis
\begin{align*}
e_{a,1}(x),\dots,e_{a,k}(x)\in E_x
\end{align*}
by
\begin{align*}
\Psi_a(e_{a,r}(x))=(x,\varepsilon_r)
\end{align*}
for $1\le r\le k$, where $\varepsilon_1,\dots,\varepsilon_k$ is the standard ordered basis of $\mathbb R^k$. We declare an ordered basis of $E_x$ to be positive if it has the same orientation as $e_{a,1}(x),\dots,e_{a,k}(x)$ for one, hence every, $a$ with $x\in V_a$.
To check independence of $a$, let $x\in V_a\cap V_b$. By definition of the transition map,
\begin{align*}
\Psi_a\circ \Psi_b^{-1}(x,v)=(x,h_{ab}(x)v).
\end{align*}
Thus the change-of-basis matrix from the frame $e_{b,1}(x),\dots,e_{b,k}(x)$ to the frame $e_{a,1}(x),\dots,e_{a,k}(x)$ is $h_{ab}(x)$. Since $\det h_{ab}(x)>0$, these two ordered bases determine the same orientation of $E_x$. Therefore the above rule gives a well-defined orientation on every fibre.
The local frames $e_{a,1},\dots,e_{a,k}:V_a\to E$ vary smoothly because they are obtained from the smooth trivialization $\Psi_a^{-1}$ applied to constant coordinate vectors. Hence the fibre orientations are locally represented by smooth positive frames. This is precisely an orientation of the real vector bundle $E$.[/step]