They've done a bit of reuse of variables here. I'm using x and X to reduce confusion.
Basically, they have taken an array, with separation h between each point. Each mass is initially at equilibrium. Let us consider a ball at a distance x from the beginning of the grid, and solve the problem in its vicinity. Now the equilibrium positions of the particles in the neighborhood are x, x+h, x+2h, etc. Now, they have defined u(X) as the displacement from equilibrium of the mass whose equilibrium position is X. So, u(x) is the displacement of particle at x, u(x+h) is displacement of particle at (x+h), and so on. Then they've used this notation go derive stuff.
The reason that they've chosen a particle at x and solved for its neighborhood, instead of taking the particle at the leftmost end, is that the final wave equation must be a function of x and t. (more accurately, $x-vt$). So then they get a general expression for displacement of particle at x (after limiting h to zero, we get a continuous row of particles, instead of a discrete one, so the notation of x becomes more appropriate)
Note that actually there is no leftmost end, as $N\to\infty$.