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.
This is true for any smooth circle action on a closed manifold. There is no need to have an almost complex structure.
In fact, more generally we have the following.
Suppose $S^1$ acts smoothly on a closed manifold $M$. Then the fixed point set $M^{S^1}$ has the property that $\chi(M) = \chi(M^{S^1})$.
(So, in particular, the result holds even if the fixed point set is not isolated).
Proof: By averaging an arbitrary Riemannian metric on $M$, we may assume the action is isometric. The fixed point set of an isometric action is always a totally geodesic submanifold, but it may have several components. (It can only have finitely many components since it is compact). In particular, asking about $\chi(M^{S^1})$ makes sense.
We also point out that in this situation, the number of isotropy groups is finite. (This is true, up to conjugacy, for any compact Lie group action on any closed manifold, as a consequence of the slice theorem).
Let $N$ be a component of $M^{S^1}$ and let $\nu N$ be an embedding of the normal bundle of $N$ into some $\epsilon$-neighborhood of $N$. By shrinking $\epsilon$, we may assume that $\nu N_1 \cap \nu N_2 = \emptyset$ for disjoint components $N_1,N_2\subseteq M^{S^1}$. I will use the notation $\nu M^{S^1}$ to denote the the union of the $\nu N_i$
Note that the $S^1$ action preseves $\nu M^{S^1}$ since is is characterized as the set of points a distance $< \epsilon$ away from $M^{S^1}$ and the action is isometric.
It follows that the $S^1$ action also preserves $M\setminus \nu M^{S^1}$. Since we've removed the points with isotropy group $S^1$, and all other closed subgroups of $S^1$ are finite (and there are only finitely many of them) there is a neighborhood $U$ of the identity $1\in S^1$ with the property that any $p\in U$ acts with no fixed points. In particular, the action field ( $\frac{d}{dt}|_{t=0} e^{it} \cdot m$, for $m\in M$) has no zeros. By Poincare-Hopf, $M\setminus \nu M^{S^1}$ has zero Euler characteristic.
In a similar fashion, the $S^1$ action preserves the boundary $\partial \nu M^{S^1}$ since $\partial \nu M^{S^1}$ consists of all points in $M$ a distance $\epsilon$ from $N$. Repeating the argument in the previous paragraph, we deduce $\partial \nu N$ also has zero Euler characteristic.
Now we are basically done. Write $M = (M\setminus \nu M^{S^1}) \cup \nu M^{S^1}$. Using the fact that each $\nu N_i$ deformation retracts to $N_i$ (so, in particular, $\chi(\nu M^{S^1}) = \chi(M^{S^1})$, we compute \begin{align*} \chi(M) &= \chi(M\setminus \nu M^{S^1}) + \chi(\nu M^{S^1}) - \chi(\partial \nu M^{S^1})\\&= 0 + \chi(\nu M^{S^1}) + 0 \\ &= \chi(M^{S^1}). \end{align*}
Best Answer
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.