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:
$$\require{AMScd}
\begin{CD}
\Omega B\times F @>>> \Omega B\\
@VVV @VVV \\
F @>>> \ast.\\
\end{CD}$$
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.)