Let $f:E\to B$ be a map of based spaces, and let $F$ be the homotopy fiber. Here is another way of constructing the action of $\Omega B$ on $F$. By definition, there is a homotopy pullback square $$\require{AMScd} \begin{CD} F @>>> \ast\\ @VVV @VVV \\ E @>>> B.\\ \end{CD}$$ Taking homotopy pullbacks along the inclusion $\ast\to B$ produces a map to the above homotopy pullback square from the following one: $$\require{AMScd} \begin{CD} \Omega B\times F @>>> B\\ @VVV @VVV \\ F @>>> \ast.\\ \end{CD}$$$$\require{AMScd} \begin{CD} \Omega B\times F @>>> \Omega B\\ @VVV @VVV \\ F @>>> \ast.\\ \end{CD}$$ The two morphisms in this square are the projections. The action of $\Omega B$ on $F$ is just the map between the top left corners of these squares; let's call this map $\mu$. This action is not just the projection: this construction shows that there is a homotopy pullback square $$\require{AMScd} \begin{CD} \Omega B\times F @>{\mathrm{pr}}>> F\\ @V{\mu}VV @VVV \\ F @>>> E;\\ \end{CD}$$ if $\mu$ was just projection onto $F$, then the space $E$ in the bottom right corner would have to be replaced by $F\times B$. (Note that this diagram shows that the composite $\Omega B \times F\to F\to E$ is trivial on $\Omega B$. You can also see this by the explicit model of this map in spaces: this composite just sends a pair $(\gamma, [e\in E, p:\ast\to f(e)])$ to $e$.) I don't know of general methods to show that the action is trivial. (Remark: one potential advantage of phrasing the construction in this way is that it works in any ($\infty$-)category with finite homotopy limits.)
