Hatcher's Theorem 3.26 on orientable of closed connected $n$-manifolds Part (a) of the theorem is as follows:
Let $M$ be a closed connected $n$-manifold. Then
(a) If $M$ is $R$-orientable, the map $H_n(M;R) \rightarrow H_n(M|x;R) \approx R$ is an isomorphism for all $x \in M$.
I am wondering what the map he is referring to. I tried to look at the proof to deduce what the map is but he uses some abstract lemma that I don't really comprehend (to be honest I have found the entire section on orientations and homology to be vague and nearly unreadable due to lack of details).
I THINK he is referring to the quotient map from the long exact sequence of the pair $(M,M-x)$ but this is only my best guess. 
 A: It's just a lot of unwinding definitions... if $A$ is a subset of $X$ then we can look at the group of "relative cocycles" $C_n(X,A;R):=C_n(X;R)/C_n(A;R)$. Then you check that the boundary maps $\partial:C_n(X;R)\to C_{n-1}(X;R)$ send $C_n(A;R)\to C_{n-1}(A;R)$, so $\partial$ induces a well-defined map $\partial:C_n(X,A;R)\to C_{n-1}(X,A;R)$.
Now, when taking the homology $H_n(X;R)=Z_n(X;R)/B_n(X;R)$, you check that under the "natural map" $C_n(X;R)\to C_n(X,A;R)$, elements of $B_n(X;R)$ are sent to $B_n(X,A;R):=\operatorname{im}\partial\subseteq C_n(X,A;R)$ (and similarly for elements of $Z_n$) so there is a well-defined natural map on homology $H_n(X;R)\to H_n(X,A;R)$.
Now, the case in question is just taking $X=M$ and $A=M-\{x\}$. If you're trying to write out this map $H_n(M;R)\to H_n(M|x;R)$ explicitly, you just need to follow the "procedure" outlined above and be careful. It is definitely tedious, but once you write it out and see what's going on it'll begin to feel more natural.
