fibered category associated to a pseudo functor If necessary for other users, I will write down the definitions for pseudo functor and/or fibered category, otherwise they can find them in page $42,44$   of Angelo vistoli's part Grothendieck topologies, fibered categories and descent theory in book FGA explained.
Given a fibered category $\mathcal{F}\rightarrow \mathcal{C}$ with a cleavage fixed, we associate a pseudo functor $\Phi$ on $\mathcal{C}$. I understand this. 
It says given any pseudo functor $\Phi$ on $\mathcal{C}$ we can associate a fibered category, with a cleavage.
He does that assuming $\Phi$ is actually a functor $\Phi:\mathcal{C}^{op}\rightarrow (Cat)$ to category of categories. 
The idea is to define a category $\mathcal{F}$ with a functor $\mathcal{F}\rightarrow \mathcal{C}$ such that the fiber $\mathcal{F}(U)$ is canonically equivalent to the category $\Phi(U)$.
It says an element of $\mathcal{F}$ is a pair $(\xi,U)$ where $U$ is an object of $\mathcal{C}$ and $\xi\in \Phi(U)$. So, objects are pairs.
Let $(\xi,U),(\eta,V)$ be two objects in $\mathcal{F}$. We have to define what are the morphisms $(a,f):(\xi,U)\rightarrow (\eta,V)$ between these. We expect morphisms to be pairs as objects are also pairs. Choice for $f$ is more or less obvious i.e., a morphism $f:U\rightarrow V$ in $\mathcal{C}$. This $f$ gives map $\Phi(f):\Phi(V)\rightarrow \Phi(U)$ and $\eta\in \Phi(V)$ mapsto an element $\Phi(f)(\eta)\in \Phi(U)$. So, we have two elements $\xi,\Phi(f)(\eta)$ and we pick a morphism $\xi\rightarrow \Phi(f)(\eta)$ and call it $a$.  Thus, morphism set is also clear.
Now, we need to define what does it mean to say composition between two morphisms. Coniser $$(\xi,U)\xrightarrow{(a,f)} (\eta,V)\xrightarrow{(b,g)}(\zeta,W)$$
This composition should give a map $(\xi,U)\xrightarrow{(c,h)} (\zeta,W)$ and one   obvious choice for $h:U\rightarrow W$ is just the composition $g\circ f:U\rightarrow V\rightarrow W$ and $c$ is expected to be morphism $$\xi\rightarrow \Phi(g\circ f)(\zeta)$$
that is an arrow in $\Phi(U)$.
But author writes something else for $c$. He writes
$$(b,g)\circ (a,f)=(\Phi(f)(b)\circ a,g\circ f)$$ 
This is not clear for me. 
We have $a:\xi\rightarrow \Phi(f)(\eta)$. $\Phi(f):\Phi(V)\rightarrow \Phi(U)$. 
As $b:\eta\rightarrow \Phi(g)(\zeta)$ is an arrow in $\Phi(V)$, $\Phi(f)(b)$ gives an arrpw in $\Phi(U)$ we have $$\Phi(f)(b):\Phi(f)(\eta)\rightarrow \Phi(f)(\Phi(g)(\zeta))=\Phi(g\circ f)(\zeta)$$ (This is because $\Phi$ is a functor).
So, we have $a:\xi\rightarrow \Phi(f)(\eta)$ in $\Phi(U)$ and $\Phi(f)(b):\Phi(f)(\eta)\rightarrow \Phi(g\circ f)(\zeta)$ and their composition 
$$\Phi(f)(b)\circ a:\xi\rightarrow \Phi(g\circ f)(\zeta)$$ 
which is the same thing as I have expected but what I do not understand is what is the point of writing down as $\Phi(f)(b)\circ a$ when you can just write $\xi\rightarrow\Phi(g\circ f)(\zeta)$?
Is it just because we can not write $\Phi(g\circ f)=\Phi(f)\circ\Phi(g)$ if $\Phi$ is just a pseudo functor or Is there any other reason that I am missing? 
 A: I think, I understand what your problem is. To start off, the definition of the composition he gives is the correct one. Assume that someone starts with a functor $\Phi: \mathsf{C}^{{\rm op}} \to  \mathsf{Cat}$. Then this functor-"like" object conists of a data, which fulfils some compatibility conditions (you can find them in his notes somewhere). Assume now that you have constructed the corresponding fibered category over $\mathsf{C}$, say $\mathsf{F}$. Pick out three arbitrary objects, $(\xi, V), (\eta, W),(\zeta,U)$ and think of the following diagram
$$(\xi,U)\xrightarrow{(a,f)} (\eta,V)\xrightarrow{(b,g)}(\zeta,W),$$
where the pairs of morphisms conist of the following: $f:U \to V$, $g:=V \to W$, are morphisms in $\mathsf{C}$, while $a: \xi \to \Phi(f)(\eta)$, $b: \eta \to \Phi(g)(\zeta)$, are mophisms in $\Phi(U)$ and $\Phi(V)$ correspondingly. Now since the composition (whatever it is) must give morphisms of the same form, we should end up with a morphism $(\tau,h): (\xi, U) \to (\zeta, W)$. Since clearly $h=g \circ f$,the second morphism must be something like $\tau:\xi \to \Phi(g \circ f)(\zeta)$.
We define now the composition to be $(b,g) \circ (a,f)=(\Phi(f)(b) \circ a, 
g \circ f)$. As I said, the composition $g \circ f$ is obvious. Regarding its first part, notice that $\Phi(f): \Phi(V) \to \Phi(U)$ from the definition of pseudo-functor is a functor and $b$ is a morphism in the category $\Phi(V)$, therefore we obtain $\Phi(f)(b):\Phi(f)(\eta) \to \Phi(f)(\Phi(g))(\zeta)=\Phi(g \circ f)(\zeta)$. So defining $\Phi(f)(b) \circ a$ gives a morphism in $\Phi(U)$, hence our choice is well-defined. Clearly this morphism has domain $\xi$, and codomain $\Phi(g \circ f)(\zeta)$. Thus this definition of composition is totally consistent with the definition of a morphism in $\mathsf{F}$. 
