What part of the fundamental group is captured by the second homology group?

$H_2(X)$ is all about $\pi_1(X)$ and $\pi_2(X)$. If $\pi_2(X)$ is trivial (as for knot complements) then it is a functor of $\pi_1(X)$.

Let $H_n(G)$ be $H_n(BG)$, the homology of the classifying space ($K(G,1)$). If $X$ is path-connected than there is a surjection $H_2(X)\to H_2(\pi_1(X))$ whose kernel is a quotient of $\pi_2(X)$, the cokernel of a map from $H_3(\pi_1(X))$ to the largest quotient of $\pi_2(X)$ on which the canonical action of $\pi_1(X)$ becomes trivial.

This $H_2(G)$ isn't anything like the next piece of the derived series after $H_1(G)=G^{ab}$, though. For example, if $G$ is abelian then $H_2(G)$ is the second exterior power of $H_1(G)$ (EDIT: so it can be nontrivial even though it knows no more than $H_1(G)$ does), while if $H_1(G)$ is trivial $H_2(G)$ is often nontrivial (EDIT: so, even when it does carry some more information than $H_1(G)$, it is not necessarily derived-series information).

EDIT: The previous paragraph comes from looking at the integral homology Serre spectral sequence of $X\to K(\pi_1(X),1)$, where the homotopy fiber is the universal cover $\tilde X$. Since $H_1\tilde X=0$, the groups $E^\infty_{p,1}$ are trivial and we get an exact sequence $$ 0\to E^\infty_{0,2}\to H_2(X)\to E^\infty_{2,0}\to 0, $$ therefore $$ E^2_{3,0} \to E^2_{0,2}\to H_2(X)\to E^2_{2,0}\to 0. $$ Since $H_2(\tilde X)=\pi_2(\tilde X)=\pi_2(X)$, this looks like $$ H_3(\pi_1(X)) \to H_0(\pi_1(X);\pi_2(X))\to H_2(X)\to H_2(\pi_1(X))\to 0. $$ The place to look for the rest of the derived series would be homology with nontrivial coefficients, for example homology of covering spaces.

A slight expansion on my comment, sort of complimentary to Tom's response.

In complete generatlity $H_2X$ tells you nothing about $\pi_1 X$.

If $X = A \times B$ with $A$ a $K(\pi,1)$ and $B$ a $K(\pi,2)$, provided $H_2(A)=0$, you have that $H_2 X = H_2 B$.

Since there are lots of $K(\pi,1)$ spaces with $H_2$ trivial, this allows you to construct many spaces with identical fundamental groups yet $H_2$ varies wildly.

You'll want to restrict to fairly particular spaces to avoid this independence.

edit: If you're happy taking covering spaces then $H_2$ (of of an arbitrary cover of $X$) starts to see quite a bit more of $\pi_1 X$. If $\widetilde{X} \to X$ is the universal cover then $H_2 \widetilde{X} \simeq \pi_2 X$ by the Hurewicz theorem. So now Tom's comments apply, giving you a concrete relationship between $H_2 X$, $\pi_1 X$ and $\pi_2 X = H_2 \widetilde{X}$.

If $X$ is a $K(G,1)$, then the homology is that of the fundamental group, and you'd have Hopf's formula.

If $G = F/R$ where $F$ is a free group, then the formula says that $H_2(G,\mathbb{Z}) \cong (R \cap [F,F])/[F,R]$.