Tangent bundle $TM\to M$ is an orientable bundle iff $M$ is orientable This is Example 6.3 in Bott-Tu, which asserts a smooth manifold $M$ is orientable iff the tangent bundle $TM\to M$ is an orientable bundle.
If $A=\{(U_\alpha,\psi_\alpha)\}$ is an atlas for $M$, then for each $\alpha$, there is a local trivialization $\phi_\alpha:TU_\alpha\to U_\alpha \times \Bbb R^n$ (where $n=\dim M$) given by $\sum_{i=1}^n a^i \dfrac{\partial }{\partial x^i}|_p$ where $\psi_\alpha=(x^1,\dots,x^n)$. Clearly the transition function $g_{\alpha \beta}:U_\alpha\cap U_\beta \to GL_n(\Bbb R)$ equals the Jacobian $U_\alpha\cap U_\beta \to GL_n(\Bbb R)$, $p\mapsto J(\psi_\alpha \circ \psi_\beta^{-1})(p)$. Thus if $A$ is an oriented atlas, then the trivialization $\{(U_\alpha, \phi_\alpha)\}$ is oriented, and this proves one direction.
But how does the opposite direction hold? (There is no explanation in the book)
 A: The answer is given by Kajelad in the comment
You have proved that $M$ is orientable $\Rightarrow TM$ is orientable. Now we prove the opposite direction.
Assume that $TM$ is orientable. Then there is a family of open cover $\{U_i\}_{i\in \Lambda}$ of $M$ and for each $i\in \Lambda$, a local trivialization
$$\varphi_i : TU_i \to U_i \times \mathbb R^n$$
so that for all $i, j$ with $U_i \cap U_j \neq \emptyset$, the transition function
$$ g_{ij} : U_i \cap U_j \to \operatorname{GL}_n(\mathbb R)$$
has $\det g_{ij} >0$.
By shrinking to smaller open sets if necessary, we assume that each $U_i$ is a coordinates neighborhood. That is, there is $\psi_i : U_i \to \psi (U_i) \subset \mathbb R^n$ which is a local chart. By composing with a reflection of $\mathbb R^n$ if necessary, we assume that in $U_i$, both
$$\left\{ \frac{\partial }{\partial x_1}, \cdots, \frac{\partial }{\partial x_n}\right\}, \{ \varphi^{-1}_i e_1, \cdots, \varphi^{-1}_i e_n\}$$
have the same orientation. Here $\{e_1, \cdots, e_n\}$ are the standard basis of $\mathbb R^n$. This is the same as saying that for all $x\in U$, the linear map $L_i (x)$ defined by the composition
$$ \mathbb R^n \cong T_{\psi(x)} \psi_i (U_i))\overset{(\psi^{-1}_i)_*}{\to} T_xU \overset{\varphi_i|_{T_xU_i}}{\to} \mathbb R^n$$
has positive determinant.
Now we check that $M$ is orientable: whenever $U_i \cap U_j$ is non-empty, let $x\in U_i$. Then one need to check $J_{ij}:= J(\psi_i \circ \psi_j):\mathbb R^n \to \mathbb R^n$ has positive determinant, but this is true since
\begin{align}
J(\psi_i \circ \psi_j^{-1}) &= (\psi_i)_* \circ (\psi_j^{-1})_* \\
&= (L_i^{-1} \circ \varphi_i) \circ (\varphi_j^{-1} \circ L_j)\\
&= L_i^{-1} \circ g_{ji} \circ L_j. 
\end{align}
