$\require{AMScd}$ 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 horizontal maps form homotopy fiber sequences and 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 the left square is a map $\Omega F_g \to ?$, which is an equivalence since the square is homotopy cartesian. On 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.