Assume that $M$ is an $n$-dimensional smooth manifold. Although I have seen many threads and online handouts dealing with these two different ideas I cannot understand what makes them equivalent. Why the choice of continuous local orientation implies smooth orientation for $M$? The definitions I am using are the following
$\mathbf{1.)}$ Smooth orientation for a manifold $M$ is a maximal atlas $\{ (U_{\alpha}, \phi_{\alpha}) \}_{\alpha \in A}$, such that the transition map $\phi_{\alpha} \circ \phi^{-1}_{\beta}$, $\forall \alpha, \beta \in A$, has positive Jacobian determinant.
$\mathbf{2.)}$ Local (homological) orientation for a smooth manifold $M$ is the choice of generator $[M]_x \in H_n(M,M-\{x\}) \cong \mathbb{Z}$, $\forall x \in M$. We call such a choice continuous, if for every $x \in M$, there exists a neighbourhood $x \in U \subset M$ and a class $a \in H_n(M,M-U)$, such that the inclusion $(M, M-U) \to (M,M-\{x\})$, induces a homomorphism $H_n(M,M-\{x\}) \to H_n(M,M-U)$ which sends $a$ to $[M]_x$. A continuous local orientation for $M$ is called homological orientation.
Best Answer
$2 \to 1$
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).