An orientation of a manifold can be defined in various ways. Let's suppose all throughout that $M$ is a connected smooth manifold (possibly with boundary), of dimension $m$. If $\omega, \eta \in \Omega^{m}(M)$ are two non-vanishing top forms on $M$, then one can prove that there exists a non-vanishing $f \in C^{\infty}(M)$ such that $\omega = f \eta$. Since $M$ is connected and $f$ is non-vanishing, it is either positive everywhere or negative everywhere. If $f$ is positive everywhere we say $\omega$ and $\eta$ define the same orientation. In this definition, an orientation is an equivalence class of non-vanishing top forms (where the equivalence relation is "define the same orientation" as defined above). An oriented manifold is then a manifold $M$ along with a choice of orientation, i.e. a particular choice of a non-vanishing top form.
We usually say that $S^2$ is an orientable manifold, meaning it can be given an orientation. You would consider $S^2$ an oriented manifold if you have fix an orientation of it. One usually takes $\omega = x\ dy \wedge dz - y\ dx \wedge dz + z\ dx \wedge dy$ as the orientation for $S^2$.
It is important in Stokes' theorem to understand that if you have an oriented manifold with boundary, it induces an orientation on the boundary in a natural way. The most intuitive way that I know to explain this is the following: suppose $M$ is your connected manifold with boundary and $\omega$ is your choice of non-vanishing top form on $M$. First we fix $X$ a non-vanishing outward-pointing vector field on $\partial M$. By outward-pointing I mean that if we take coordinates $x^1, \ldots, x^m$ near a point in the boundary, and the boundary is defined by the condition $x^m = 0$, then the vector field $X$ has an everywhere negative $\frac{\partial}{\partial x^m}$ component. Such a vector field can always be obtained (construct it locally then patch it together by partitions of unity). Then we define a non-vanishing top form on $\partial M$ given by $(\iota_X \omega)(X_1, \ldots, X_{m - 1}) = \omega(X, X_1, \ldots, X_{m - 1})$. It turns out that this choice of orientation doesn't depend on the choice of $X$, as long as it is outward-pointing (for a proof of this, see Lee's Introduction to Smooth Manifolds, in the chapter about orientations). Hence this orientation on $\partial M$ depends only on the choice of orientation we had in $M$. This is the choice of orientation we always make for the boundary of an oriented manifold.
It is important to note one thing about the induced boundary orientation. Set $\mathbb{H}^{m} = \{(x_1, \ldots, x_m) \in \mathbb{R}^{m}\ |\ x_m = 0\}$. Note that $\partial \mathbb{H}^{m} \cong \mathbb{R}^{m - 1}$. The induced boundary orientation on $\partial \mathbb{H}^{m}$ matches with the natural orientation of $\mathbb{R}^{m - 1}$ when $m$ is even, but is the opposite orientation when $m$ is odd.
You are correct in your belief that the oriented hypothesis here is crucial. In general, we can only integrate forms on an oriented manifold. For a non-oriented manifold, there is a more general concept of density.
For any top-form $\omega$, $$\int_M L_X(\omega) = \int_{M} d(i_X(\omega)) + i_X(d\omega) = 0, $$ as by Stokes' theorem, the integral of any exact top-form on a manifold without boundary is zero, and because $d\omega$ is zero, as it is an $(n+1)$-form on an $n$-dimensional manifold.
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Integration on an oriented manifold is defined only for compactly supported differential forms $\alpha$, and Stokes theorem is also stated for compactly supported one. If $\alpha$ is a compactly supported differential form on $M$, $\alpha|_{\operatorname{Int}M}$ might not be of compact support on $\operatorname{Int} M$. So in general you cannot apply Stokes theorem to $\alpha|_{\operatorname{Int}M}$ (There are cases where integration are defined and Stokes theorem are applicable for more general differential forms, but definitely not in full generality).