Why are higher homology groups not the abelianizations of higher homotopy groups? Really the question is exactly the title:

Why (conceptually and geometrically if possible) are higher homology groups not the abelianizations of higher homotopy groups?

 A: There's a paper Daniel M. Kan, A combinatorial definition of homotopy groups, Ann. of Math. (2) 67 (1958), 282–312s, that actually shows rather nicely that in certain contexts, the higher homology groups ARE abelianizations of higher homotopy groups. The Hurewicz theorem also gives a context in which this is true. 
The paper. "Chapter III" describes "in precisely what sense the homology groups are abelianized homotopy groups," according to the third paragraph of the introduction. 
A: By triangulating the sphere, there's a map $\pi_n(X,x_0) \to H_n(X;\Bbb Z)$ for all $n$ (after noting that homotopic maps are sent to homologous cycles). This is known as the Hurewicz homomorphism. The theorem you want is 

The Hurewicz homomorphism for $n=1$ is surjective with kernel $[\pi_1(X,x_0),\pi_1(X,x_0)]$.

Surjectivity is easy here, because every $1$-cycle can easily be represented by a (homologous) map $S^1 \to X$. Our first problem with generalizing is that there's no obvious way to represent an $n$-cycle as a map from the $n$-sphere; indeed, $H_2(T^2) \cong \Bbb Z$, but $\pi_2(T^2) = 0$. (To see the latter, note that because $S^2$ is simply connected, you can lift a map $S^2 \to T^2$ to its universal cover $\Bbb R^2$, which is contractible. Now project this null-homotopy back down.) 
Even if you have surjectivity, there's the problem of calculating the kernel. I would say that this is possible precisely because 1-cycles are so simple and there's not much data to carry around; you may disagree. Either way, I suggest reading the actual proof; you can find it e.g. in Hatcher's algebraic topology book, theorem 2A.1.
All this is what makes the following theorem (the Hurewicz theorem) sort of miraculous.

Let $X$ be simply connected and let $m$ be the least integer such that $\pi_m(X)$ is nonzero. Then $H_k(X)=0$ for $k<m$ and the Hurewicz homomorphism for $n=m$ is an isomorphism.

A: If you have a closed n-dimensional submanifold $M\subset X$, then you can triangulate it and get a cycle that represents an element in $H_n(X;{\mathbb Z})$. (It might be trivial, and in higher dimensions not all homology classes of a manifold are obtains this way, but never mind, it's just an example.)
If $M$ was an embedded n-sphere, then it also represents an element in $\pi_nX$. (I.e., the homology class is in the image of the Hurewicz homomorphism.) But in general, if $M$ is not a sphere, then the homology class will not be represented by an element of the homotopy group.
For example, consider a Torus or a surface of higher genus. By triangulating it you get an element in the second homology. But this element will not be representable by a sphere, so $\pi_2\to H_2$ is not surjective in this case.
