How much is cohomotopy dual to homotopy?

The homotopy groups can be written as covariant homotopy invariant functors $\pi_n:\mathrm{Top}_\ast\to\mathrm{Set}$. If we were to consider contravariant homotopy invariant functors $\pi^n:\mathrm{Top}_\ast^{op}\to\mathrm{Set}$, we would obtain the cohomotopy sets. How dual is it? Well, $\pi^n(S^m)=\pi_m(S^n)$. If $X$ is a CW-complex of dimension (at most) $n$, then $\pi^p(X)\to H^p(X)$ is a bijection. See the nlab. As for whether or not a "cohomotopy extension property" exists, I don't know; it seems like an interesting thing!

All homology/cohomology is with integer coefficients.

The Hurewicz map that you are looking is defined as follows I think: Given a homotopy class $f:X\rightarrow S^n$, and $[u]\in H^n(S^n)$ you get $f^*[u]\in H^n[X]$. If you take now $[u]$ to be the positive generator of $H^n(S^n)\cong \mathbb{Z}$ you get a nice map $h:\pi^n\rightarrow H^n(X)$.

If $X$ is an $n$-dimensional CW complex any map into an $n+k$-connected space (e.g. n+k dim sphere) is homotopically trivial. Combining this with the observation that $pi^n\cong H^n(X)$ for $n$-dimensional closed manifolds we see that $\pi^l(X)\cong H^l(X)$ for all $l\geq n$ for $n$ dimensional closed manifolds. Furthermore $\pi^0(X)$ measures the number of connected components of $X$ and as $S^1$ is a $K(\mathbb{Z},1)$ we have that $\pi^1(X)\cong H^1(X)$.