@Josh: I don't have any precise reference in mind for that, maybe in the Coddington-Levinson? but as far as I remember it is mostly for linear ODE's

Note first that reversing time $t\to -t$ is equivalent to changing $\sigma\to-\sigma$, so you only need to study one side (say $t\to+\infty$). The case $\sigma=0$ is a borderline one that I am not too sure how to deal with. I assume below that $\sigma\neq 0$ (but I guess $\sigma$ is the propagation speed of the wave, so it should be OK to discard stationary waves).

The starting point is to rewrite $w''+\sigma w'+f(w)=0$ as
$$
(e^{\sigma t}w')'=-e^{\sigma t}f(w).\hspace{2cm}(E)
$$

**Step 1**
The first thing you need to show is that $w_{\pm}$ are necessarily steady-states, i-e $f(w_{\pm})=0$ (stationary equilibrium solutions of the ODE, usually one is stable while the other is unstable). In order to see this assume by contradiction that $f(w_+)\neq 0$ (again, it is enough to look at $t\to+\infty$).

- If $\sigma>0$ then by (E) we have $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ not integrable when $t\to\infty$, so $e^{\sigma t}w'\sim C\int e^{\sigma t}=Ce^{\sigma t}$ hence $w'\sim C\neq 0$. This shows that $w$ blows-up linearly and contradicts $w(\infty)=w_+$.
- If now $\sigma<0$ then $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ becomes integrable, and thus $e^{\sigma t}w'\to C$ for some limit $C\in \mathbb{R}$. If $C\neq 0$ then $w'\sim Ce^{-\sigma t}$ blows exponentially, which contradicts again $w(\infty)=w_+$. Thus $C=0$, and integrating $(e^{\sigma t}w')'\sim Ce^{\sigma t}$ from $t$ to $\infty$ you get $e^{\sigma t}w'-0\sim C(e^{\sigma t}-0)$, hence again linear blow-up $w'\sim C\neq 0$.

**Step 2**
Once you know that $f(w)\to f(w_+)=0$ (here I am definitely using the continuity of $f$) the heuristic idea is quite simple: the initial ODE $w''+\sigma w+f(w)=0$ roughly becomes a 1st order linear ODE in $v=w'$
\begin{equation}
v'+\sigma v=-f(w)\approx -f(w_+)= 0,\qquad v=w'.\hspace{2cm}(E')
\end{equation}

This linear ODE $v'+\sigma v=0$ gives either the trivial solution $v'=v=0$ (which means precisely $w''=w'=0$), or $v'$ and $v$ proportional to $e^{-\sigma t}$. If $\sigma>0$ you see that both the trivial and exponential cases are admissible and lead to $v',v=w'',w'\to 0$ when $t\to\infty$. If now $\sigma<0$ the exponential blow-up $v=w'\sim e^{-\sigma t}$ is excluded because you assume $w(t)\to w^+=cst$, so the only possibility is again $w'',w'=u',u=0$.

Of course this step 2 is only formal, and rigorously justifying (E') from (E) requires tedious and technical computations similar to those in step 1. For that you may want to use again (E) with now $f(w)\to f(w_+)=0$ hence
$$
(e^{\sigma t}w')'=o\left(e^{\sigma t}\right),
$$
and distinguish again integrability or linear/exponential blowup at infinity as in step 1.

I hope this helps!