Let $K$ be a knot in $S^3$, and N(K) be its tubular neighborhood. I want to compute the homology of $S^3-N(K)$ using Mayer-Vietoris. Let $A$ be $N(K)$ minus a small neighborhood, and $B$ be the complement of N(K) with a small neighborhood added. Note that $A\cup B$ is $S^3$, and $A\cap B$ deformations retracts onto a torus. $H_1(S^3-N(K))$ and $H_n(S^3-N(K))$, $n>2$, pop out of Mayer Vietoris applied to A and B. However I am struggling with $H_2$. According to Mayer-Vietoris we have the following exact sequence $$H_3(S^3) \to H_2(T) \to H_2(A)\bigoplus H_2(B) \to 0$$ The part I don't understand is the map $H_3(S^3) \to H_2(T)$. Let $\alpha$ be the generator of $H_3(S^3)$. Under the boundary map of Mayer-Vietoris, $\alpha$ is subdivided into a part residing in $A$ and a part residing in $B$. Then we take the boundary of the part residing in $A$. Thinking of the boundary map like this makes me think that it is an isomorphism. However, I am not sure of this.
Are there other ways to see what this map does to the generator of $H_3(S^3)$? More generally, how does one work with the boundary map in the MV sequence in higher dimensions, where things cannot be simply visualized?
You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = \partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} \setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$\partial U = - \partial V = [T]$.
Then, by definition, the boundary map maps $U+V \in H_{3}(S^{3},\mathbb{Z})$ to $\partial U \in H_{2}(T,\mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.