Optimal Loading of a Hopping Airplane Minimum Cost Flow Problem
One approach to understand some complicated situation is to figure out some simplest but still meaningful situations.

In the current situation, let us check a situation when there are only two hops, hop 0 and hop 1. The simpler situation when there is only one hop is too trivial to be instructive, sine there will be no passenger to pick up.

Picking up your pen (or pencil or stylus or your finger), draw the following reduced version of Figure 9.4 in the reference provided.

 

The above direct network means 


there are up to $b_{12}$ passengers at node 1-2 who can be picked up at node 1 (hop 1) and who should be dropped at node 2 (hop 2)  
The cost of the arc from node 1-2 to node 1 is $-f_{12}$, where $f_{12}$ is the fare per passenger from node 1 to node 2. That means every passenger picked up will pay $f_{12}$.
The capacity of arc from node 1 to node 2 is $p$. There is no associated cost.
The arc from node 1-2 to node 2 is the imaginary flight with no cost associated nor any capacity limit.
There should be $b_{12}$ passengers reaching node 2. They either take the flight from node 1 to node 2 physically or take the arc from node 1-2 to node 2 virtually without paying any fare. 


The minimum cost flow problem for this simple directed network is the following, where $x_{a,b}$ means the number of passengers (the flow) on the arc from $a$ to $b$.


  $$\text{Minimize }\ \ z(x)= (-f_{1-2, 1})x_{1-2,1} + 0x_{1,2} + 0x_{1-2, 2}$$
  subject to 
  $$x_{1-2,1} + x_{1-2,2} = b_{12}$$
  $$x_{1,2} - x_{1-2,1} = 0$$
  $$-x_{1-2,2} - x_{1,2} = -b_{12}$$
  $$ 0\le x_{1-2, 1}$$ 
  $$ 0\le x_{1, 2}\le p$$ 
  $$ 0\le x_{1-2,2}$$


Why do we want to minimize $z(x)$? Because the total fare collected is $-z(x)$. 

All the conditions in the end just means, $x_{1-2,1}\le b_{12}$ and $x_{1-2,1}\le p$. We can solve this problem minimum cost flow easily by letting $x_{1-2,1}=\min(b_{12},p)$, the most passengers available not exceeding the airplane capacity.

Hopefully the above explanation is detailed enough to enable you to further understand the full minimum cost flow formulation for the problem of optimally loading a hopping airplane. You can establish a one-to-one correspondence between feasible passenger routings and feasible flows in the minimum cost flow formulation of the problem.