$\require{AMScd}$
Here are some details. Without loss in generality, we can assume $f$ and $g$ are fibrations. I am assuming that all spaces are based. We may assume that $F_f$ now refers to the actual fiber of $f$ (which is equivalent to its homotopy fiber).


Start with the commutative diagram
$$
\begin{CD}
F_f @>>> A @> f >> B\\
@VVV @VVV @VV gV \\
F_{g\circ f} @>>> A @>> g\circ f > C
\end{CD}
$$
Now induce once over to the left to get a diagram
$$
\begin{CD}
\Omega B @>>> F_f @>>> A \\
@VVV @VVV @VVV \\
\Omega C @>>> F_{g\circ f} @>>> A 
\end{CD}
$$
where the left square is homotopy cartesian (reason: the map horizontal fibers
of this square is identified with the identity map of $\Omega A$).


The vertical fibers of this square is a map $\Omega F_g \to ?$, which is an equivalence since the square is homotopy cartesian.
One the other hand the square fits into a diagram
\begin{CD}
\Omega B @>>> F_f \\
@VVV @VVV  \\
\Omega C @>>> F_{g\circ f}\\
@VVV @VVV \\
F_{g} @= F_{g}
\end{CD}
It follows that there's a homotopy fiber sequence
$$
\Omega F_g \to F_f \to F_{g\circ f} 
$$
and the map $\Omega C \to F_g$ factors through $F_{gf}$.

It follows from the long exact sequence of homotopy groups that 
$F_f \to F_{g\circ f} \to F_g$ is a homotopy fiber sequence.