For now, let $M$ be any closed oriented $n$-dimensional manifold, and $f : S^m \to M$ a smooth map.

First suppose $k \geq 1$.

Since $m = n + 4k > 2$, we have $H^2(S^m; \mathbb{Z}_2) = 0$, and hence $f^*w_2(TM) = 0 = w_2(TS^m)$, so every map $f : S^m \to M$ is a spin map.

If $\omega$ is a volume form on $M$ with $\int_M\omega = 1$, then the $\hat{A}$-degree of $f$ is $\int_{S^n}f^*\omega\wedge\hat{A}_k(TS^m) = 0$ since $\hat{A}_k(TS^m)$ is a polynomial in the Pontryagin classes of $TS^m$ which vanish as $TS^m$ is stably trivial.

Now suppose $k = 0$, so $m = n$.

If $m \neq 2$, then the same argument applies to show that every map $f : S^m \to M$ is spin. If $m = 2$, then $M$ is a closed oriented surface and hence spin, so $f^*w_2(TM) = f^*0 = 0 = w_2(TS^2)$; again, we see that every map is a spin map.

When $k = 0$, the $\hat{A}$-degree of $f$ is precisely the usual degree of $f$. There exists a non-zero degree map $f : S^m \to M$ if and only if $\widetilde{M}$, the universal cover of $M$, is a rational homology sphere, see [here](https://math.stackexchange.com/q/2271532/39599). There are examples of such manifolds with non-negative curvature operator, such as the Wu manifold $SU(3)/SO(3)$.