If $M$ and $N$ are spin, then every map between them is a spin map. In particular, there exist spin maps $M \to N$ of degree zero.
If $M$ is spin and $N$ is not spin, then $f : M \to N$ is a spin map if and only if $f^*w_2(TN) = 0$. So nullhomotopic maps are spin maps of degree zero for example.
If $M$ is not spin and $N$ is spin, then there are no spin maps $f : M \to N$ (the condition to be satisfied is $f^*0 = w_2(TM) \neq 0$).
If $M$ and $N$ are both not spin, there may or may not be spin maps between them, and even if there are, there may be none of degree zero.
Let $M = \mathbb{CP}^2$ and $N = (S^2\times S^2)/\mathbb{Z}_2$ where $\mathbb{Z}_2$ acts by the antipodal map on both factors. Both $M$ and $N$ are non-spin, but there are no spin maps between them. To see this, note that any map $f : \mathbb{CP}^2 \to (S^2\times S^2)/\mathbb{Z}_2$ lifts through the universal covering $\pi : S^2\times S^2 \to (S^2\times S^2)/\mathbb{Z}_2$ to a map $\tilde{f} : \mathbb{CP}^2 \to S^2\times S^2$ since $\mathbb{CP}^2$ is simply connected. Since $f = \pi\circ\tilde{f}$ and $S^2\times S^2$ is spin, we see that $$f^*w_2(TN) = \tilde{f}^*\pi^*w_2(TN) = \tilde{f}^*w_2(\pi^*TN) = \tilde{f}^*w_2(T(S^2\times S^2)) = \tilde{f}^*0 = 0 \neq w_2(TM).$$
If $M = N = \mathbb{CP}^2$, then a map $f : M \to N$ is a spin map if and only if it has odd degree. In particular, there are no spin maps of degree zero. For an example with $M \neq N$, let $M = \mathbb{CP}^2\#(S^2\times S^2)$ and $N = \mathbb{CP}^2$, then the degree one map $M \to N$ is a spin map. One can also show there are no spin maps of degree zero here either. More generally, we have the following:
Let $M$ be a closed, connected, oriented four-manifold. Then there is a spin map $f: M \to \mathbb{CP}^2$. Moreover, there is a spin map $f : M \to \mathbb{CP}^2$ of degree zero if and only if $w_2(TM)$ admits an integral lift $c$ with $c^2 = 0$.
Proof: Note that $[M, \mathbb{CP}^2] \cong [M, \mathbb{CP}^{\infty}] \cong H^2(M; \mathbb{Z})$. Under this correspondence, $[f] \mapsto f^*\alpha$ where $\alpha$ is a generator of $H^2(\mathbb{CP}^2; \mathbb{Z})$ . Since $\alpha$ reduces mod $2$ to $w_2(T\mathbb{CP}^2)$, we see that $f^*\alpha$ reduces mod $2$ to $f^*w_2(T\mathbb{CP}^2)$.
Note that $M$ is spin$^c$, so there is $c \in H^2(M; \mathbb{Z})$ whose mod $2$ reduction is $w_2(TM)$. Choosing an $f$ such that $f^*\alpha = c$, we see that $f^*w_2(T\mathbb{CP}^2) = w_2(TM)$, so $f$ is a spin map.
If $\beta$ denotes an oriented generator of $H^4(M; \mathbb{Z})$, then $(f^*\alpha)^2 = f^*\alpha^2 = (\deg f)\beta$, so $f$ has degree zero if and only if $(f^*\alpha)^2 = 0$. If $f$ is a spin map, then $c = f^*\alpha$ is an integral lift of $w_2(TM)$ and if $f$ has degree zero, then $c^2 = 0$. Conversely, if $c$ is an integral lift of $w_2(TM)$ with $c^2 = 0$, then the corresponding map (unique up to homotopy) $f : M \to \mathbb{CP}^2$ is a spin map of degree zero. $\square$
So, for example, $M = (S^2\times S^2)/\mathbb{Z}_2$ admits a spin map $M \to \mathbb{CP}^2$ of degree zero because $w_2(TM) \neq 0$ has integral lift the unique non-zero element $c$ of $H^2(M; \mathbb{Z}) \cong \mathbb{Z}_2$ which necessarily satisfies $c^2 = 0$.
A corollary of the above result is that if there is a spin map $f : M \to \mathbb{CP}^2$ of degree zero, then $w_2(TM)^2 = 0$ (as is the case for $M = (S^2\times S^2)/\mathbb{Z}_2)$.