If $X$ is a differentiable manifold, so that both notions are defined, then they coincide.
The ``patching'' of local orientations that you describe can be expressed more formally as follows: there is a locally constant sheaf $\omega_R$ of $R$-modules on $X$ whose stalk at a point is $H^n(X,X\setminus\{x\}; R).$ Of course, $\omega_R = R\otimes_{\mathbb Z} \omega_{\mathbb Z}$.
This sheaf is called the orientation sheaf, and appears in the formulation of Poincare duality for not-necessarily orientable manifolds. It is not the case that any section of this sheaf gives an orientation. (For example, we always have the zero section.)
I think the usual definition would be something like a section which generates each stalk.
I will now work just with $\mathbb Z$ coefficients, and write $\omega = \omega_{\mathbb Z}$.
Since the stalks of $\omega$ are free of rank one over $\mathbb Z$, to patch them together you
end up giving a 1-cocyle with values in $GL_1({\mathbb Z}) = \{\pm 1\}.$ Thus underlying
$\omega$ there is a more elemental sheaf, a locally constant sheaf that is a principal bundle for $\{\pm 1\}$. Equivalently, such a thing is just a degree two (not necessarily connected) covering space
of $X$, and it is precisely the orientation double cover of $X$.
Now giving a section of $\omega$ that generates each stalk, i.e. giving an orientation of $X$, is precisely the same as giving a section of the orientation double cover (and so $X$ is orientable, i.e. admits an orientation, precisely when the orientation double cover is disconnected).
Instead of cutting down from a locally constant rank 1 sheaf over $\mathbb Z$ to just a double cover, we could also build up to get some bigger sheaves.
For example, there is the sheaf $\mathcal{C}_X^{\infty}$ of smooth functions on $X$.
We can form the tensor product $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega,$
to get a locally free sheaf of rank one over ${\mathcal C}^{\infty}$, or equivalently, the sheaf of sections of a line bundle on $X$. This is precisely the line bundle of top-dimensional forms on $X$.
If we give a section of $\omega$ giving rise to an orientation of $X$, call it $\sigma$, then we certainly get a nowhere-zero section
of $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega$, namely $1\otimes\sigma$.
On the other hand, if we have a nowhere zero section of $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z}
\omega$, then locally (say on the the members of some cover $\{U_i\}$ of $X$ by open balls) it has the form $f_i\otimes\sigma_i,$ where $f_i$ is a nowhere zero real-valued function on $U_i$ and $\sigma_i$ is a generator of $\omega_{| U_i}.$
Since $f_i$ is nowhere zero, it is either always positive or always negative; write
$\epsilon_i$ to denote its sign. It is then easy to see that sections $\epsilon_i\sigma_i$
of $\omega$ glue together to give a section $\sigma$ of $X$ that provides an orientation.
One also sees that two different nowhere-zero volume forms will give rise to the same orientation if and only if their ratio is an everywhere positive function.
This reconciles the two notions.
I doubt that in general one can construct a reasonable sheaf on $U$ with the required properties. To see what kind of bad things can happen, let us try to understand why this works for $X$ an elliptic curve and the sheaf $\cal{O}^{\times}$ on it.
We have the derived global sections functor from the $D^b$ of sheaves on $X$ to the $D^b$ of sheaves on a point, i.e. graded vector spaces. But given a sheaf $F$ on $X$ we can compute its global sections in a roundabout way: we can first take the pullback to $U$, then take global sections and then take the $G$-invariants where $G=\pi_1(X)$. Passing to the derived categories we get $$R\Gamma (F)=R(R\Gamma f^{-1}(F))^G$$ where $F\in D^b(X)$, $f:U\to X$ is the projection, $R\Gamma f^{-1}$ is the right derived functor of the left exact functor $\Gamma f^{-1}$ and $R(\cdot)^G$ is the right derived functor of the functor of $G$-invariants (this functor goes from the $D^b$ of $G$-modules to graded vector spaces).
We have the Grothendieck spectral sequence that converges to $H^\ast(X,F)$ with the $E_2$ sheet given by $$E_2^{p,q}=H^p(G,H^q(U,f^{-1}(F)).$$
Now if $X$ is an elliptic curve, $F=\cal{O}^{\times}$ and $U=\mathbf{C}$, then it follows from the exponential exact sequence that $H^q(U,f^{-1}(F))=0$ for $q\neq 0$, so the above spectral sequence collapses and we get the required isomorphism $H^\ast (X,F)=H^\ast(G,H^0(U,f^{-1}(F))$. This also happens when say $F$ is locally constant and $U$ is contractible. But in general there seems no reason to expect the spectral sequence to collapse, let alone to be concentrated in one row only.
Best Answer
An elaboration.
$\mathcal O_n$ is a bundle over $M$ with fiber $\mathbb Z$. There is an action of the integers on it, because the integers act on homology. Earlier I said this was a principal bundle, I was too tired! The action is of course not free. In particular, this bundle $\mathcal O_n \to M$ has a section.
Given a manifold $M$, its orientation cover is a $\mathbb Z_2$-principal bundle. The action of $\mathbb Z_2$ is given by reversal of orientations.
So the product $\tilde M \times \mathbb Z$ has an action of $\mathbb Z_2$ given by
$$t.(x,n) = (t.x, (-1)^tn)$$
where $t.x$ is the action of $\mathbb Z_2$ on $\tilde M$.
$\tilde M \times_{\mathbb Z_2} \mathbb Z$ is the quotient of $\tilde M \times \mathbb Z$ by the action of $\mathbb Z_2$. By design, it is a $\mathbb Z$-principal bundle -- you can project onto $\tilde M / \mathbb Z_2 \equiv M$ -- and by design it is isomorphic to $\mathcal O_n$. The map $\tilde M \times_{\mathbb Z_2} \mathbb Z \to \mathcal O_n$ is induced by your inclusion $\tilde M \to \mathcal O_n$.