$R$ orientation in Hatchers The precise source is Chapter 3, page 235, - line 9.
So what Hatcher has stated is:

$M$ is an $n$ manifold, and $R$ is a ring, and $x \in M$. Then
$H_n(M,M-x;R) \simeq H_n(M,M-x; \Bbb Z) \otimes_{\Bbb Z} R$

Why is this true? This is not clear $- \otimes_{\Bbb Z} R:Ab \rightarrow Ab$ is not in general exact.
 A: You are right that it's not exact in general, however you can use the universal coefficient theorem and a computation of $H_{n-1}(M,M-x; \mathbb{Z})$ to prove that it is in fact correct. 
A: By the universal coefficient theorem we have the short exact sequence:
$$0\to H_i(X, A,\mathbb{Z})\otimes R\to H_i(X,A,R)\to Tor(H_{i-1}(X, A,\mathbb{Z}), R)\to 0$$
So to obtain the isomorphism we want $Tor$ to vanish and it is enough to show that $H_{i-1}(X,A,\mathbb{Z})$ is free (or more generally flat).
So in your case all you have to know is that $H_{n-1}(M, M-\{x\},\mathbb{Z})$ is free. The homology calculation can be found in Hatcher page 231 but let me repeat the argument:
$$\begin{matrix}H_i(M,M-\{x\}) & \simeq & H_i(\mathbb{R}^n,\mathbb{R}^n-\{0\}) &\text{by excision} \\
& \simeq & \tilde{H}_{i-1}(\mathbb{R}^n-\{0\}) &\text{by contractibility} \\
& \simeq & \tilde{H}_{i-1}(S^{n-1}) & \text{by homotopy equivalence} \\
\end{matrix}$$
which is $\mathbb{Z}$ if $i=n$ and $0$ otherwise. In any case it is free and so the isomorphism holds for any $i$, not only $n$.
Finally $M$ being a manifold is crucial. Generally $Tor$ does not have to vanish. Probably even for $(X,X-\{x\})$ pairs the relative homology might not be flat (over $\mathbb{Z}$ flat is equivalent to free) and so there will be $R$ with $Tor$ non-zero, although I don't have a concrete example.
