This is a lemma from Analysis on manifolds by Munkres page 317:
Let $M$ be a compact $n$-manifold in $\mathbb{R}^n$, oriented naturally. Let $\omega=h\,dx_1\wedge \cdots\wedge dx_n$ be an $n$-form defined in an open set of $\mathbb{R}^n$ containing $M$. Then $h$ is the corresponding scalar field, and $$\int_M \omega=\int_M h\,\mathrm{d}V.$$
Proof is the following:
we have $$\int_M \omega=\int_M \lambda\, \mathrm{d}V,$$ where $\lambda$ is obtained by evaluating $\omega$ on an orthonormal basis for $T_p(M)$ that belongs to its natural orientation. Now $\alpha$ belongs to the orientation of $M$ if det $D\alpha>0$; thus the natural orientation of $T_p(M)$ consists of the right-handed frames. The usual basis for $T_p(M)=T_p(\mathbb{R}^n)$ is one such frame, and the value of $\omega$ on this frame is $h$.
My problem: The natural orientation of $T_p(M)$, induced by the orientation of $M$ due to the textbook is the collection of all $k$-frames in $T_p(M)$ of the form $$((\alpha(\mathbf{x});D\alpha(\mathbf{x})\cdot\mathbf{a}_1),\ldots,(\alpha(\mathbf{x});D\alpha(\mathbf{x})\cdot\mathbf{a}_n))$$ where $(\mathbf{a}_1,\cdots,\mathbf{a}_n)$ is a right-handed frame in $\mathbb{R}^n$ and $\alpha : U\rightarrow V$ is a coordinate patch belonging to the orientation of $M$ about $\alpha(\mathbf{x})=\mathbf{p}$ . In the proof How det $D\alpha>0$ reduces this collection to just right-handed frames ?
the usual basis for $T_p(M)$ consists of $(\mathbf{p};D\alpha(\mathbf{x})\cdot\mathbf{e}_j)$ and usual basis for $T_p(\mathbb{R}^n)$ consists of $(\mathbf{p};\mathbf{e}_j)$ but these two linear spaces being equal how one should get rid of $\alpha$?