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.
One way to go about this is to start with your manifold $M \subset \mathbb R^{m+1}$ and consider a height function $f : \mathbb R^{m+1} \to \mathbb R$ (orthogonal projection onto a vector) restricted to $M$. So there is some fixed vector $v \in S^m$ such that $f(x)=\langle x,v\rangle$ for al $x \in \mathbb R^{m+1}$.
Generically, this is a Morse function so its gradient (the orthogonal projection of $v$ to $TM$) is a vector field which is transverse to the $0$-section of $TM$. So Poincare-Hopf tells you how you can compute the Euler characteristic from this.
Now how is that related to the Gauss map? If you were to compute the degree of the Gauss map $\nu : M \to S^m$ by computing its intersection number with $v \in S^m$ you would have a very similar looking sum to compute! But do notice that the orthogonal projection of $v$ to $TM$ can be zero at both $\nu^{-1}(v)$ and $\nu^{-1}(-v)$. When you work out the details this ultimately explains why there's the $1/2$ and why it only works in even dimensions.
I hope that gives you the idea without giving too much away.
Best Answer
Consider the following vector field $\vec v$ on $S^2$. Using standard spherical coordinates $\theta$, $\phi$, and let $\hat u_\theta$ and $\hat u_\phi$ be the unit vectors in the $\theta$ and $\phi$ directions, $\vec v$ is given by:
$$ \vec v = \begin{cases} v_0\sin(2\theta)\hat u_\theta +v_0\sin(\theta)\hat u_\phi &\text{if $\theta\ne0$},\\ 0 & \text{if $\theta=0$}, \end{cases}$$
where $v_0$ is a positive constant. This vector field has only two zeros (at the N and S poles) and both zeros are sources (if you make $v_0<0$ then they are both sinks).