I was recently reading some texts on Thom classes and the Thom isomorphism (e.g. Differential Forms in Algebraic Topology by Bott and Tu). There it is stated that one of the defining characteristics of the Thom class of a rank-$n$ vector bundle $\pi:E\rightarrow M$ is that it restricts to a generator of the cohomology group $H^n_c(F_x)\cong H^n(S^n)$ for all $x\in M$, where I used the fact that compactly supported cohomology is isomorphic to ordinary cohomology of the one-point compactification.
Now, on the other hand some authors state the theorem in terms of $H^n(\mathbb{R}^n, \mathbb{R}^n\backslash\{0\})$ which is isomorphic to $H^{n-1}(S^{n-1})$ by a long exact sequence. As such state the existence of a Thom class in terms of fundamental classes and orientations (which makes it useful for a generalization to other orientation with respect to generalized cohomology theory). In fact, Hatcher states both formulations.
Obviously, on the level of groups, the two paragraphs are equivalent since both groups are $\mathbb{Z}$. But I fail to see the cohomological/geometrical equivalence except for the fact that the pair $(\mathbb{R}^n, \mathbb{R}^n\backslash\{0\})\approx (D^n, \partial D^n)$ can give rise to either of the groups (one by a long exact sequence as already stated and the other through the quotient $D^n/\partial D^n$. Intuitively this means that an orientation of the sphere $S^{n-1}$ is equivalent to an orientation of the sphere $S^n$ due to an inclusion of the equator (I guess?)
Is this last paragraph the right way to obtain a relation between the different formulations? (My apologies if the formulation of my question is a bit confusing. This would be due to my own confusion.)