I think I have found a more elementary approach to the problem, so I'll post it for anyone who might be interested (and maybe to check whether I haven't made some silly mistake).
The idea is actually quite simple: I approximate the flow to the first order and use this to get a lower bound on the periods of nonfixed points.
Proposition: Let $X$ be a smooth vector field on $\mathbb R^n$ such that $|X|$ and $|dX|$ are bounded. Then there is a $\tau >0$ such that for all $0<t<\tau$: $$\theta(t,p) = p \quad \iff\quad X(p) = 0$$
Proof: By Taylor expansion we have
$$\theta(t,p) = p + tX(p) + \int_0^t (t-\tau) \; dX\left(\theta\left(\tau,p\right)\right) X\left(\theta\left(\tau,p\right)\right) \; d\tau$$
By choosing $t_0$ small enough, we may assume
$$\left|X\left(\theta\left(\tau,p\right)\right)\right| \le 2\left|X(p)\right|$$
for all $p\in \mathbb R^n$ and $0\le \tau \le t < t_0$.
Edit: As has been pointed out by David Speyer in the comments, the existence of such a $t_0$ isn't as clear as I had initially thought. To see that such $t_0$ exists, we assume $|dX|<M$ for some $M>0$ and $|X| < \tilde M$. By Taylorexpansion we have
$$\left|X\left(\theta\left(\tau,p\right)\right)\right| \le |X(p)| + \tau M |X\left(\theta\left(s,p\right)\right)|$$
where $s \in [0,\tau]$ is chosen to maximize $|X\left(\theta\left(s,p\right)\right)|$. Now let $t_0 := 1/(2M)$. By iterating the same argument with $|X\left(\theta\left(s,p\right)\right)|$ we get the estimate
\begin{align*}
\left|X\left(\theta\left(\tau,p\right)\right)\right| &\le |X(p)| + \tau M \; \big(\; |X(p)| + \tau M |X\left(\theta\left(s,p\right)\right)|\; \big) \\
&\vdots \\
&\le \sum_{k=0}^\infty \left(\tau M\right)^k |X(p)| + \lim_{k\to \infty} (\tau M)^k\tilde M \\
&\le 2|X(p)|
\end{align*}
Now let us define $$\Phi(t,p) = p + t X(p)$$
From the above and the properties of $X$, there is some $C>0$ and $t_0>0$ such that for $0<t<t_0$
$$|\theta(t,p) - \Phi(t,p) | = \left|\int_0^t (t-\tau) \; dX\left(\theta\left(\tau,p\right)\right) X\left(\theta\left(\tau,p\right)\right) \; d\tau\right| \le C|X(p)|t^2$$
for all $p$. But then
\begin{align*}
|\theta(t,p) - \theta(0,p)| &\ge |\Phi(t,p) - \theta(0,p)| - |\theta(t,p) - \Phi(t,p)| \\
&\ge t|X(p)| - t^2C|X(p)| \\
&= t|X(p)| (1 - Ct)
\end{align*}
So if $p$ is a point such that $\theta(T,p) = \theta(0,p)=p$ it follows that either $X(p)=0$ or $T \ge C^{-1}$. Proving the proposition.
Hint: For $p\in X$, let $U_p$, with coordinates $(x_1,...,x_n,t)$, be a slice chart around $p$ (meaning around $p$, $X$ corresponds to points where $t=0$).
Now, given your normal vector field $V$, orient $X\cap U_p$ by declaring the ordered basis $\{\partial_{x_i}\}$ to be positively oriented iff the ordered basis $\{\partial_{x_i}, v\}$ is positively oriented in $Y$.
Conversely, if $X$ is oriented, define $V = \partial_t$.
I'll leave it to you to prove that all this works.
Best Answer
Here's a fun proof that employs the Transversality theorem to show that on any smooth manifold $M$, there is a vector field that vanishes only on a $0$-dimensional submanifold of $M$. Of course, when $M$ is compact, every $0$-dimensional submanifold is finite, which gives you your desired result.
Assume without loss of generality that $M^n$ is embedded in $\mathbb{R}^N$ with $N>n$. Define a map $F:X\times \mathbb{R}^N\to TX$ by $F(p,v)=\text{proj}_{T_pM}v$. Then $F$ is a smooth submersion. In particular, $F$ is transverse to $Z=X\times \{0\}$. So, by the transversality theorem, there exists some $v\in \mathbb{R}^N$ so that $f_v(x):X\to TX$ is transverse to $Z$. Now, $f_v(x)$ is a smooth section of $TX$, and so $f_v$ is a vector field. So $f_v^{-1}(X\times \{0\})$-the zeros of $f_v$-is a submanifold of $X$ of codimension $\dim TX-\dim X\times \{0\}= \dim X$, as claimed.