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.
This is an application of the inverse function theorem. Define
$$ \Phi(x^1,\dots,x^m) = (F^1(x^1,\dots,x^m),x^2,\dots,x^m) $$
and denote the coordinates in the codomain by $u,v^2,\dots,v^m$. The differential of $\Phi$ at $\mathbf{a} = (a^1,\dots,a^m)$ is
$$ d\Phi|_{(a^1,\dots,a^m)} = \begin{pmatrix} \frac{\partial F}{\partial x^1}(\mathbf{a}) & \frac{\partial F}{\partial x^2}(\mathbf{a}) & \dots & \frac{\partial F}{\partial x^m}(\mathbf{a}) \\
0 & 1 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1\end{pmatrix} $$
and so since $\frac{\partial F}{\partial x^1}(\mathbf{a}) \neq 0$, the differential is invertible at $a$ so you can invert $\Phi$ locally near $a$. Call the inverse
$$G = G(u,v^2,\dots,v^m) = G(u,\mathbf{v}) = \left(G^1(u,\mathbf{v}), \dots G^m(u,\mathbf{v}) \right). $$ Then
$$ (\Phi \circ G)(u,\mathbf{v}) =
\left( F^1(G^1(u,\mathbf{v}),\dots,G^m(u,\mathbf{v})), G^2(u,\mathbf{v}),\dots,G^m(u,\mathbf{v}) \right)
= \\
(u,v_2,\dots,v_m). $$
This implies that
$$ (F \circ G)(u,\mathbf{v}) = \\
\left( F^1(G^1(u,\mathbf{v}), \dots, G^m(u,\mathbf{v})), F^2(G^1(u,\mathbf{v}), \dots, G^m(u,\mathbf{v})),\dots, F^n(G^1(u,\mathbf{v}),\dots,G^n(u,\mathbf{v})) \right) \\
=
\left( u, F^2(G^1(u,\mathbf{v}),\mathbf{v}), \dots F^n(G^1(u,\mathbf{v}),\mathbf{v}) \right).$$
So if we abuse notation and write $F(u,\mathbf{v})$ for $F \circ G$ and similarly for the coordinates of $F$ then we have
$$ F(u,v_1,\dots,v_m) = \left( u, F^2(u,\mathbf{v}), \dots, F^n (u,\mathbf{v}) \right). $$
Best Answer
Well, $e^{i\theta (z)}=z\implies \theta(z)=\ln z/i$.
Note, your $S^1$ sits in $\Bbb C$.
Now, we need a branch of log, because the complex logarithm is not a single valued function. The principal branch can be assumed, for instance, in which case, we have $\ln z=\ln r +i\theta$, where $z=re^{i\theta}$.
In this case your $c$ will be $0$. For a different branch of log, it will be $\ln z=\ln r+i\theta+2\pi ki$. See https://math.stackexchange.com/a/1475592/403337.
So let's compose and see what we get: $\hat\epsilon(t)=\theta\circ\epsilon(t)=\theta(e^{2\pi i t})=\ln(e^{2\pi i t})/i=2\pi t+2\pi k =2\pi t+ c$, where $c=2\pi k $.