There is a theorem that says that the normal bundle of the boundary a manifold is trivial, hence orientable
It goes like this :
Let $n=dim M$. Let $\mathbb{R}^n_+=\{x\in \mathbb{R}^n : x_1\geq 0\}$. And let $\pi :\mathbb{R}^n \rightarrow \mathbb{R}$ be the projection $\pi(x)=x_1$. Let $\{\phi_i:U_i\rightarrow \mathbb{R}^n_{+}\}$ be a family of charts of $M$ that cover $\partial M$.We can define morphisms $F_i : TM|\partial U_i \rightarrow, F_ix=D(\pi\circ \phi_i)_x$, and we will have that these induce morphisms $G_i:v|\partial U_i\rightarrow \mathbb{R}$ which will be a bimorphism .Note that the transition functions of these linear maps have positive determinant and this proves that $v$ is orientable.Then the rest of the proof is just to take partitions of unity and I have been able to check that in fact we have an isomorphism to the trivial bundle. My question why is $G_{jx}\circ G_{ix}^{-1}:\mathbb{R}\rightarrow \mathbb{R}$ positive ? I have tried to see what this map is and I think it's the following, assuming $\phi_1=(x_1,...,x_n)$ and $\phi_j=(y_1,...,y_n)$ we have that the map sends $a $ to $[a \frac{\partial }{\partial x_1}]$ and then this is the same as $[\frac{y_1}{x_1}(x)a\frac{\partial }{\partial y_1}]$ and this goes $\frac{y_1}{x_1}(x)a$, and so I guess we are saying that $\frac{y_1}{x_1}(x)>0$ but why is this the case ? Why can't it be negative, it probably has to due with the fact that we are working in $\mathbb{R}^n_+$ but I don't see why it's true . Any clarification of this is appreciated, Thanks.