How can we apply generalized Stokes' theorem to a non-oriented manifold with boundary? I do not really know much about the boundary of non-oriented manifold.
A boundary of oriented manifold, if it exists, has a sign. If you reverse the orientation, the boundary picks up an extra negative sign.
Then, how can we talk about a boundary of non-oriented manifold, as it won't pick up a negative sign, if you reverse the orientation?
Sorry, I have a very rudimentary knowledge on homology, and de Rham cohomology.
 A: In (co)homology, a typical way of generalizing a statement concerning orientable objects is to replace the (co)homology theory with a "twisted" version that reduces to the usual version in the orientable case. In our situation, the ability to integrate a global form depends on having an orientation, but with twisted cohomology we can create a kind of "twisted orientation".
(Warning: my exposition is a bit sloppy, especially with respect to compactly supported forms on non-compact manifolds. Please consider my answer as a map to topics to learn more about, rather than a definitive source.)
For de Rham cohomology, there is a description in Bott and Tu's "Differential Forms in Algebraic Topology", Chapter 7. For a vector bundle $E$ of rank $n$, its $n$-th alternating power $\Lambda^n E$ is a line bundle, called the "determinant line bundle of $E$", and $E$ is orientable iff $\Lambda^n E$ has a nowhere-vanishing section, iff $\Lambda^nE$ is a trivial bundle. For a smooth manifold $M^n$, Bott and Tu denote $\Lambda^n TM$ by $L$, and let 
$$ \Omega^q(M, L) = \Gamma((\Lambda^qT^*M)\otimes L)$$
Why does this help? Well, if $M$ is closed and orientable, we normally prove $H^n_{dR}(M) \cong \mathbb{R}$ by constructing a volume form in $\Omega^n(M)$ and then showing that integration along this form induces the above isomorphism. But a volume form is precisely a nowhere-vanishing section of $\Lambda^n T^*M$. If $M$ is non-orientable then its cotangent bundle is not orientable, so $\Lambda^nT^*M$ doesn't admit a nowhere-vanishing section; however, 
$$\Lambda^nT^*M \otimes L = \Lambda^nT^*M \otimes \Lambda^nTM \cong Hom(\Lambda^nTM,\Lambda^nTM)$$
does admit such a section, namely the section which is $id$ in every fibre. So non-orientable manifolds admit a "twisted volume form" aka a density which can be used to show $H^n_{dR}(M,L)\cong \mathbb{R}$ for any closed $n$-manifold. The "intuitive" idea is that we can't define a global volume form because of the way the tangent bundle twists, but we sort-of can if the coefficients twist in a compatible way.
The old theory is recovered in the orientable case, since $L$ is a line-bundle that admits a nowhere-vanishing section and therefore is trivial, so $E\otimes L \cong E$ for any vector bundle $E$ over $M$ and so $\Omega^q(M, L) \cong \Omega^q(M)$.
This twisted theory can be used (as in Bott and Tu) to generalize the Thom Isomorphism to non-orientable bundles, and to generalize Poincare Duality and Stokes' Theorem to non-orientable manifolds. Their specific statement for Stokes' Theorem is the following (Theorem 7.7):

On any manifold $M$ of dimension $n$, orientable or not, if $\omega\in \Omega^{n-1}_c(M,L)$, then $$ \int_M d\omega = 0. $$

