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)
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}