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Johannes Ebert
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There is a construction for both Thom isomorphisms, homological and cohomological, via classical stable homotopy theory. You find the details in Rudyaks book "On Thom spectra, orientability, and cobordism", chapter V, §1. The Thom class is a map $X^{\mu} \to\Sigma^{n} H \mathbb{Z}$. Moreover, there is a map of spectra $X^{\mu} \to X_+ \wedge X^{\mu}$ which is induced from the map of vector bundles $\mu \to \mathbb{R}^0 \times \mu$ over the diagonal map $X \to X \times X$. Here is the definition of the homological Thom isomorphism; the cohomological one is in the same spirit. Consider the composition

$X^{\mu} \wedge H \mathbb{Z} \to X_+ \wedge X^{\mu} \wedge H\mathbb{Z} \to X_+ \wedge \Sigma^n H \mathbb{Z} \wedge H \mathbb{Z} \to X_+ \wedge \Sigma^n H \mathbb{Z} $. On homotopy groups, it induces a map lowering the degree by $n$ (there is a sign mistake in your question that confused me for some minutes).

It is clear that this works for orientations with respect to other ring spectra as well.

Johannes Ebert
  • 20.9k
  • 4
  • 74
  • 117