Let $p\in M$, let $f\in C^\infty\left(N\right)$, and let $X\in T_pM$. By assumption, $\left(F_*X\right)\left(f\right)=X\left(f\circ F\right)=0$. Let $\left(U,\varphi\right)$ be a smooth chart containing $p$. Then
$$X=\sum_iX^i\left.\frac{\partial}{\partial x^i}\right|_p=\sum_iX^i\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)},$$
which implies that
$$\left(\sum_iX^i\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\right)\left(f\circ F\right)=\sum_iX^i\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\left(f\circ F\circ\varphi^{-1}\right)=0,$$
which in turn, by basic calculus, implies that $F$ is constant on $U$, i.e., $F\left(x\right)=c$ for some $c\in N$ and every $x\in U$.
Since $M$ is connected, it is the case that $M$ is path connected. Let $q\in M$ and let $\gamma:\left[0,1\right]\to M$ be a path connecting $p$ and $q$. Since, as above, $F$ is constant on each smooth chart $\left(U_{\gamma\left(x\right)},\varphi_{\gamma\left(x\right)}\right)$ containing $\gamma\left(x\right)$ for every $x\in\left[0,1\right]$, it is the case that $F\equiv c$ on $M$ since $F\left(p\right)=c$ and $\gamma$ is continuous.
The degree indeed depends on the orientations of the manifolds $M$ and $N$: If you change the orientation of either $M$ or $N$ (or both) the degree changes sign (or keeps the same, respectively).
So one should better say: The degree is defined if $M$ and $N$ are oriented (rather than just orientable). It is just laziness that one speaks about orientable only, because this is of course the crucial topological property which $M$ or $N$ should have - the choice of the actual orientiation is then just a “convention” (which one should specify, of course, if one really wants to calculate a degree of a map).
Best Answer
Here's a specific map which is actually surjective. Considering $S^1$ as $\{(x, y): x^2 + y^2 = 1\}$ in the Euclidean plane, take the projection $\pi_x: \mathbb{R}^2 \to \mathbb{R}, (x, y) \mapsto x$. Define $f: \mathbb{R} \to \mathbb{R}^2, t \mapsto (\cos \pi t, \sin \pi t)$. Finally define $g$ as the composition restricted to $S^1$: $f \circ (\pi_x | S^1): S^1 \to S^1$. $\pi_x$ and $f$ are both smooth, so $g$ is too. $\pi_x$ is nullhomotopic, so $g$ is nullhomotopic and therefore has degree $0$. And it's clear that (roughly speaking) $g$ goes through all angles in the range $[-\pi, \pi]$ and is therefore onto.
Edit: fixed scaling mistake, clarified domain and range.