Let $M$ be an $n$ dimensional manifold and $R$ be a commutative ring. $\Gamma_c(S;R)$ denotes the space of sections onto the orientation bundle with compact support. For any closed set $K$, $H_n(M,M\backslash K;R)\cong \Gamma_c(K;R)$. Now, let us assume $K$ is compact and connected. If we consider the right-hand side there is no way to restrict us (at least I don't see) to take a constant section valued any element in $R$. So $\Gamma_c(K;R)\cong R$ (no matter the orientability of $K$). But on the other hand, there is also a fact: if $M$ is non-orientable along $K$ then $H_n(M,M\backslash K;R)\cong \{r\in R| r=-r\} $. Is this a contradiction?
And another question is is there any difference (or significance) to take consider an arbitrary coefficient ring in comparison to $\mathbb{Z}$?
This assumes that the orientation bundle actually has nonzero-everywhere sections. In the case where $M$ is the Mobius strip (or $\Bbb RP^2$) this assumption is false: the orientation bundle is twisted, and it doesn't even make sense to talk about a "constant section" (as it would if the orientation bundle were just $M \times R$). Which is why, taking $M = \Bbb R P ^2$ and $K = M$, we get $$ H_n(M, \emptyset; \Bbb Z) = 0 $$ which makes sense, because $\Gamma(M;R) = 0$. (Note that "compact support" here is irrelevant, because $M$ is a compact manifold to start with.)