understanding Thom's theorem for $\mathrm{MU}$ and Milnor-Novikov's result.

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Sorry if this has been asked before. $\newcommand{\MU}{\mathrm{MU}}$

I'm trying to understand the complex cobordism spectrum $\MU$, but I don't fully understand Thom's theorem, namely, that

$$ \MU_\ast\simeq\pi_\ast\MU, $$

where $\MU_\ast:=\MU_\ast(\mathrm{pt})$.

Is there a good intuition for what this isomorphism means, in the sense that there is a relation between cobordism classes and homotopy groups (why should this be true?).

If we let $R:=\mathbf{Z}[x_n : n>1,\deg x_n=2n]$, then $\MU_\ast\simeq R\simeq\pi_\ast\MU$. What exactly is the reason why $R$ has generators of only even degree? Evidently, we have that $x_n$ represents the fundamental class of $\mathbf{C}\mathrm{P}^n$, but I don't think I get the immediate connection right now.

Thanks!