I'm not well-versed in algebraic topology, so I'm not completely sure how to ask this.
If I understand correctly, the fundamental group of space $X$ at point $x$ is the group of equivalence classes of loops $l:I\rightarrow X$, $I$ is the unit interval and $l(0)=l(1)=x$, under the equivalence relation of homotopy - two loops, $l$ and $k$, are homotopic if there exists a continuous map $h:I\times I \rightarrow X$ s.t. $h(0, \cdot )=l$ and $h(1, \cdot )=k$. The group operation is concatenation of loops - just go through the first loop twice as fast and then the second loop twice as fast. The nth homotopy group is the same idea just extended to the nth dimension. If a 1-homotopy is a path (loops in this case), an n-homotopy is a continuous map from the n-cube into the space s.t. the map at each end of the cube is an (n-1)-homotopy. The nth homotopy group is the set of equivalence classes of n-homotopies under an operation which is basically a high-dimensional version of concatenation of loops whose endpoints agree.
I feel I have a much better grasp of homology groups so I won't explicate it here.
My question is whether or not there is any intuition for why there is a homomorphism from the nth homotopy group to the nth homology group. What about specifically the fact that $H_1$ is the abelianization of $\pi_1$? If there isn't intuition for either of these in general, is there a way to visualize the homotopy and homology groups of a specific space and see why there is a homomorphism from one to the other?
That there is a map is not surprising: An element of $\pi_n(X)$ is nothing but a map $S^n\rightarrow X$ satisfying some conditions. But any map induces maps in homology. More specifically there is a map $H_n(S^n;\mathbb{Z})\rightarrow H_n(X;\mathbb{Z})$. Now all these maps patch together to a map $\pi_n(X)\rightarrow H_n(X;\mathbb{Z})$. Geometrically you can think of this as follows: Choose a triangulation of $S^n$ (this generates $H_n(S^n;\mathbb{Z})$). Then the Hurewicz map puts this cycle into $X$. It turns out that this is a group homomorphism.
The homology group $H_1(X;\mathbb{Z})$ is abelian, while $\pi_1(X)$ is in general not. Hence the map $\pi_1(X)\rightarrow H_n(X;\mathbb{Z})$ must factor through the abelianization $\pi_1(X)_{\mathrm{ab}}$, i.e. we obtain a map $\pi_1(X)_{\mathrm{ab}}\rightarrow H_1(X;\mathbb{Z})$. It turns out that this is an isomorphism. Now more generally if $n>1$ the Hurewicz map $h:\pi_n(X)\rightarrow H_n(X;\mathbb{Z})$ is an isomorphism if $\pi_k(X)=0$ for $k<n$: If the low dimensional topology of $X$ is not too complicated the $n$ dimensional topology is generated by spheres. You can play around with surfaces to see that this map can't be always an isomorphism (how do you map $S^2\rightarrow \Sigma_g$ , where $\Sigma_g$ is a surface of genus $g>0$, non trivially?)