Orientations on Manifold 
This is a very basic definition of orientable and very basic example 20.5 however ı could not understand definition in an good way so ı want you to explain my green writing please :) and my example please help me ı want to learn orientation on manifold if ı could not understand in a good way this ı will not understand in a good way rest of the subject please help me 
 A: First you have to have a good notion of orientation on a vector space $V$.  Since the transition matrix from one basis to another is an invertible matrix it has non-zero determinant.  Hence, it's either positive or negative.  Two bases of $V$ determine the same orientation if the transition matrix has positive determinant.
So, an orientation on $V$ is an equivalence class of bases for $V$.  You can think of the group homomorphism
$$
\begin{align}
GL(V) \overset{\Large\epsilon}{\longrightarrow} &\{-1, +1\} \\
A \longmapsto & \frac{\det A}{|\det A|}.
\end{align}
$$
Two bases whose transition matrix lives in $GL^+(V) = \epsilon^{-1}(1)$ determine the same orientation, and two bases whose transition matrix lives in other coset $\epsilon^{-1}(-1)$ determine opposite orientations.
On a smooth manifold, each tangent space is a vector space $T_pM$ of the same dimension.  A non-vanishing top form determines a basis in each tangent space in a way that varies smoothly as $p$ varies about $M$.
It's instructive to think about a non-orientable manifold, say the open Möbius strip
$$
[0, 1] \times (0, 1) / \sim
$$
where $(0, y) \sim (1, 1 - y)$.  The left and right edges of the square are identified with a half twist.  Any consistent choice of basis at each point in the interior of the square cannot be extended to the left (= right) edge.
For example, the standard basis $\{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\}$ which is dual to the $2$-form $dx \wedge dy$ cannot be extended continuously (let alone, smoothly) to the left/right sides.  The transition matrix would be
$$
\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
$$
on one side or the other.
A: Recall that an $n$-form associates to each point $p\in M$ a map $\omega_p:(T_pM)^n\to \mathbb R$. Thus for any basis $x_1,\ldots,x_n$ of $T_pM$, we have an associated real number $\omega_p(x_1,\ldots,x_n)$, which is either positive or negative. If we let $x_1,\ldots,x_n$ vary continuously with $p$ such that $x_1,\ldots,x_n$ is always a basis for $T_pM$, then $\omega_p(x_1,\ldots,x_n)$ is never $0$ and so must have constant sign. This sign gives you a notion of the orientation of the basis $x_1,\ldots,x_n$. In the case of the $1$-manifold $\mathbb R$, there is a non-vanishing $1$-form $dx$, and we have two possibilities up to sign for bases of $T_p\mathbb R$: either $1$ or $-1$, corresponding to left or right orientation. This generalizes to $\mathbb R^n$ in the way described by Sammy.
