If you had an orientation, you'd be able to define at each point $p$ a unit vector $n_p$ normal to the strip at $p$, in a way that the map $p\mapsto n_p$ is continuous. Moreover, this map is completely determined once you fix the value of $n_p$ for some specific $p$. (You have two possibilities, this uses a tangent plane at $p$, which is definable using a $(U_\alpha,\phi_\alpha)$ that covers $p$.)
The point is that the positivity condition you wrote gives you that the normal at any $p'$ is independent of the specific $(U_{\alpha'},\phi_{\alpha'})$ you may choose to use, and path connectedness gives you the uniqueness of the map. Now you simply check that if you follow a loop around the strip, the value of $n_p$ changes sign when you return to $p$, which of course is a contradiction.
(This is just a formalization of the intuitive argument.)
$2 \to 1$
Pick a homological orientation for $M$. Fix a generator of $H_n(\Bbb R^n, \setminus \Bbb R^n \setminus \{0\})$ (equivalently, make sure you know what orientation on $\Bbb R^n$ you're using).
For a chart $\varphi_\alpha: U_\alpha \to \Bbb R^n$ with $\varphi_\alpha(x) = 0$, note that there is an induced map $(\varphi_\alpha)_*: H_n(U, U - \{x\}) \to H_n(\Bbb R^n, \Bbb R^n \setminus \{0\})$. Because $\varphi_\alpha$ is a diffeomorphism, $(\varphi_\alpha)_*$ is an isomorphism. The inclusion map $H_n(U, U - \{x\}) \to H_n(M, M - \{x\})$ is an isomorphism by the excision theorem.
We say $(U_\alpha, \varphi_\alpha)$ is an oriented chart if the composite $$H_n(M, M - \{x\}) \xrightarrow{i^{-1}_*} H_n(U, U - \{x\}) \to H_n(\Bbb R^n, \Bbb R^n \setminus \{0\})$$ sends your chosen generator of the first group to the fixed generator of the last group.
The relevance of the positive Jacobian condition is that a diffeomorphism $f: (\Bbb R^n, 0) \to (\Bbb R^n, 0)$ preserves the generator of $H_n(\Bbb R^n, \Bbb R^n \setminus \{0\})$ if and only if $\det \text{Jac}_0(f) > 0.$ In fact, if you have a smooth map $f: \Bbb R^n \to \Bbb R^n$ sending $0$ to $0$, it is homotopic to $\text{Jac}_0(f): \Bbb R^n \to \Bbb R^n$, given by setting $f_{1-t}(x) = \frac{f(tx)}{t}$ and $f_1(x) = \text{Jac}_0(f)(x)$; this is a continuous homotopy of $f$ that sends nonzero points to nonzero points as long as $f$ did the same. (If $f$ did not, restrict to a smaller open set where it did and run the same argument: remember that excision allows us to do this whenever we want.) Now we just need to see the induced maps of linear maps; because $GL_n(\Bbb R)$ has exactly two components, determined by their determinant, we just need to check the induced map of two linear maps. The identity induces the identity; a reflection induces $-1$. (Use the long exact sequence of the pair $(D^n, S^{n-1})$ and the interpretation of the map on top homology as degree in $S^{n-1}$.)
Thus the induced map on $H_n(\Bbb R^n, \Bbb R^n \setminus \{0\})$ is the sign of $\det \text{Jac}_0 f$. As long as you choose your charts to preserve the homological orientation, your transition maps will always have positive Jacobian determinant.
$1 \to 2$
Same recipe as above: this time the rule is that you choose the generator of $H_n(M, M - \{x\})$ using an oriented chart (as opposed to determining which charts are oriented using where the generator goes).
Best Answer
I will sketch a way to prove this but you will have to fill in the details. They define in page 48
In page 54 they define equivelnce of cocycles $g_{\alpha\beta}$ and $g'_{\alpha \beta}$ if there exist a maps $\lambda_\alpha:U_\alpha \to GL(n,\mathbb{R})$ such that $$ g_{\alpha \beta}=\lambda_{\alpha} g_{\alpha \beta}^{\prime} \lambda_{\beta}^{-1}\qquad \text { on } U_{\alpha} \cap U_{\beta} $$ Then they give an exercise 6.2
Recall that a refinement of an open cover $\mathcal{U}=\left(U_{i}\right)_{i \in I}$ is an open cover $\mathcal{U}^{\prime}=$ $\left(U_{\alpha}^{\prime}\right)_{\alpha \in A}$ such that there exists a map $\varphi: A \rightarrow I$ with the property $$ U_{\alpha} \subset U_{\varphi(\alpha)}, \quad \forall \alpha \in A $$ We write this $\mathcal{U}^{\prime} \prec_{\varphi}$ U. Given a gluing cocycle $g_{i j}$ subordinated to $\mathcal{U}$ then its restriction to $\mathcal{U}^{\prime}$ is the gluing cocycle $g \mid \bullet$ defined by $$ \left.g\right|_{\alpha \beta}=\left.g_{\varphi(\alpha) \varphi(\beta)}\right|_{U_{\alpha \beta}} $$ You can prove using exercise 6.2 that the bundles $\left(g_{\bullet \bullet}, u, W\right)$ and $\left(g_{\bullet \bullet}^{\prime}, \mathcal{U}^{\prime}, W\right)$ are isomorphic if and only if there exist an open cover $\mathcal{V} \prec \mathcal{U}, \mathcal{U}^{\prime}$ such that the restrictions of $g$ and $g'$ to $\mathcal{V}$ are equivalent.
Let $(U_\alpha,\psi_\alpha)_{\alpha\in A}$ and $(U_{\alpha'},\psi_{\alpha'})_{\alpha'\in A'}$ be to different covers of $M$. We can take the refinement of both cover to be given by taking the intersections of the open covers and the maps of the refinement to be $\varphi(\alpha\beta')=\alpha \in A$ and $\varphi'(\alpha\beta')=\beta' \in A'$. Now we can construct the equivalence to show that both bundles are isomorphic. To construct $\lambda_{\beta \beta'}$ use the fact that the determinant is multiplicative and
$$\psi_{\alpha}\circ \psi_{\beta}^{-1}=\psi_{\alpha}\circ \psi^{-1}_{\alpha'}\circ \psi_{\alpha'}\circ\psi^{-1}_{\beta'}\circ \psi_{\beta'}\circ \psi_{\beta}^{-1}$$ so we will get $\lambda_{\beta \beta'}=sgn J(\psi_{\beta'}\circ \psi_{\beta}^{-1})$.
for the statement "M is orientable if and only if its orientation bundle is trivial". Assume you have an orientation you can choose the transition maps to be constant $1$ and this will be equivalent to the trivial line bundle by choosing the $\lambda$'s to be constant. In the other direction you can multiply the first coordinate of the maps $\psi_{\alpha}^{-1 }$ with the $\lambda_{\alpha}$ to get that the sign will be constant as $$1=\lambda_{\alpha}\cdot \lambda_{\beta}\cdot g_{\alpha \beta}$$ by the equivalence condition.