This won't involve any geometry, but here is a model-independent description of the situation as I understand it. I will not prove anything. The very short summary is that

The infinite symmetric product and singular chains are both models of the free $\mathbb{Z}$-module spectrum on a space $X$, where $\mathbb{Z}$ is regarded as a ring spectrum, and "homotopy groups of the free $\mathbb{Z}$-module spectrum on $X$" is a model-independent definition of the (ordinary) homology groups of $X$.

First, here's the simplest version of a general definition. Let $X$ be a set and let $R$ be a commutative ring. Then the *$R$-homology* of $X$ is equivalently one of the following $R$-modules:

- the $R$-module $R[X]$ of formal $R$-linear combinations of elements of $X$,
- the direct sum $\displaystyle \bigoplus_{x \in X} R$, or equivalently the colimit of the constant diagram $X \ni x \mapsto R \in \text{Mod}(R)$,
- the value on $X$ of the left adjoint to the forgetful functor $\text{Mod}(R) \to \text{Set}$,
- the value on $X$ of the unique cocontinuous functor $\text{Set} \to \text{Mod}(R)$ sending $\text{pt}$ to $R$.

Intuitively, $R$-homology is the canonical covariant way to linearize a set $X$ into an $R$-module: above I've just given four different ways of saying "the free $R$-module on $X$."

Moreover, the forgetful functor $\text{Mod}(R) \to \text{Set}$ factors through abelian groups, and hence its left adjoint also factors through abelian groups in the other direction as the composite

$$R[-] : \text{Set} \xrightarrow{\mathbb{Z}[-]} \text{Ab} \cong \text{Mod}(\mathbb{Z}) \xrightarrow{R \otimes (-)} \text{Mod}(R)$$

which is just a fancy way of saying that we can write

$$R_0(X) \cong R[X] \cong R \otimes \mathbb{Z}[X].$$

This is "universal coefficients for sets": it says that to understand the free $R$-module on a set it suffices to understand the free $\mathbb{Z}$-module / abelian group on a set.

The last description of $R$-homology above reflects "Eilenberg-Steenrod for sets," which says that $\text{Set}$ is the free cocomplete category on a point.

Now suppose we want to linearize, not sets, but spaces, by which I mean (weak) homotopy types / $\infty$-groupoids. So let $X$ be a space and let $R$ be an $E_{\infty}$-ring spectrum. Then the *$R$-homology* of $X$ is equivalently (the homotopy groups of) one of the following $R$-module spectra, where $\text{Mod}(R)$ denotes the $(\infty, 1)$-category of $R$-module spectra (and "functor" means "$(\infty, 1)$-functor"):

- the smash product $R \wedge \Sigma^{\infty}_{+} X$, where $\Sigma^{\infty}_{+} X$ is the suspension spectrum of $X$ with a disjoint basepoint,
- the homotopy / $(\infty, 1)$-colimit of the constant diagram $X \ni x \mapsto R \in \text{Mod}(R)$,
- the value on $X$ of the $(\infty, 1)$-left adjoint to the forgetful functor $\text{Mod}(R) \to \text{Space}$,
- the value on $X$ of the unique homotopy cocontinuous functor $\text{Space} \to \text{Mod}(R)$ sending $\text{pt}$ to $R$.

Intuitively, $R$-homology is the canonical covariant way to linearize a space into an $R$-module spectrum: above I've just given four different ways of saying "the free $R$-module spectrum on $X$."

The first description above should be regarded as a direct generalization of the isomorphism $R[X] \cong R \otimes \mathbb{Z}[X]$ to spaces, except that $\mathbb{Z}$ has been replaced with the sphere spectrum $\mathbb{S}$. More precisely, the forgetful functor $\text{Mod}(R) \to \text{Space}$ factors through spectra, and hence its left adjoint also factors through spectra in the other direction as the composite

$$\text{Space} \xrightarrow{\Sigma^{\infty}_{+}(-)} \text{Sp} \cong \text{Mod}(\mathbb{S}) \xrightarrow{R \wedge (-)} \text{Mod}(R).$$

In particular $\Sigma^{\infty}_{+}$, being left adjoint to the forgetful functor from spectra to spaces, should be thought of as the "free spectrum" functor, and $R \wedge (-)$, being left adjoint to the forgetful functor from $R$-module spectra to spectra, should be thought of as the "free $R$-module spectrum (on a spectrum)" functor.

The last description of $R$-homology reflects an $(\infty, 1)$-categorical version of Eilenberg-Steenrod for spaces, which says that $\text{Space}$ is the free homotopy cocomplete $(\infty, 1)$-category on a point.

Now, at long last, ordinary homology is the homotopy groups of the free $\mathbb{Z}$-module spectrum:

$$H_{\bullet}(X, \mathbb{Z}) \cong \pi_{\bullet}(\mathbb{Z} \wedge \Sigma_{+}^{\infty} X).$$

Hopefully I've phrased things so it's clear that this story about linearizing spaces is a direct analogue of the story about linearizing sets, provided you are willing to accept (that various $(\infty, 1)$-categorical machinery works the way it ought to and) that the correct analogue of abelian groups in this setting is spectra.

Here are some more things that ought to be true and that connect this story back to more model-dependent considerations.

- By a suitable version of the stable Dold-Kan theorem, the $(\infty, 1)$-category of $\mathbb{Z}$-module spectra should be equivalent to the $(\infty, 1)$-category presented by unbounded chain complexes of $\mathbb{Z}$-modules. This should restrict to an equivalence between connective $\mathbb{Z}$-module spectra and connective chain complexes.
- By the usual Dold-Kan theorem, the category of connective chain complexes of abelian groups is equivalent to the category of simplicial abelian groups (and there should be model structures on both sides making this a Quillen equivalence presenting an equivalence of $(\infty, 1)$-categories, and so forth). This equivalence more or less sends singular chains on a topological space $X$ to the free simplicial abelian group on the singular simplicial set of $X$, and modulo technical details this gives rise to the relationship between singular homology and the homotopy groups of the free $\mathbb{Z}$-module spectrum on $X$, which is connective since $\mathbb{Z}$ and suspension spectra are connective.
- The analogue of the free simplicial abelian group on a simplicial set for topological spaces is the free topological abelian group; this is roughly what the infinite symmetric product attempts to be, and modulo technical details (in particular, niceness hypotheses on $X$) this gives rise to the relationship between the homotopy groups of the infinite symmetric product and the singular homology of $X$.

1more comment