According to p.122, Chapter II, of Bredon's Topology and Geometry:
16.1 Theorem (Thom-Pontryagin) The above construction gives an isomorphism of $\pi_{n+k}(S^n)$ with the group of framed cobordism classes of framed $k$-manifolds in $\mathbb{R}^{n+k}$.
Here and in what follows $k$ is assumed to be $k\ge 1$.
Question: Are similar results known to hold for $\pi_{n+k}(M^n)$, where $M^n$ is an arbitrary (compact) smooth $n$-manifold?
I.e. to what (known) extent are higher homotopy groups $\pi_{n+k}(M^n)$ similar to cobordism classes of $k$-manifolds?
A reference will suffice for an answer; please don't forget to include the page number.
We have (I think) for any $n$-manifold $M^n$, not just $S^n$, that $H_{n+k}(M^n)=\{0\}$ for all $k \ge 1$. Also, any smooth manifold is homotopically equivalent to a CW complex, for which all homology theories satisfying the Eilenberg-Steenrod axioms coincide.
Citing the Pontryagin-Thom construction, I would explain the intuition for the homology groups $H_{n+k}(S^n)$ vanishing to be that homology "ignores all information about cobordisms" (because if two $(n+k)$-cycles $c_1,c_2$ are such that $c_1-c_2=\partial d$ for some $(n+k+1)$-cycle $d$, then $[c_1]=[c_2]\in H_{n+k}(S^n)$). Thus, since by the Pontryagin-Thom construction the only information provided by $\pi_{n+k}(S^n)$ is about cobordisms, the Hurewicz homomorphism sends all of it to $0$.
Can we cite a similar intuitive reason for the homology groups $H_{n+k}(M^n)$ vanishing for an arbitrary smooth $n$-manifold $M^n$?