Recurrence relation for the number of "references" to two mutually recursive function
It looks like that you are exhausted after a long and tiring journey that you have taken to comprehend the setup of assembly-line scheduling, the step 1 on the structure of the fastest way through the factory and the step 2 on a recursive solution.
It is straightforward to understand formula (4) and (5).
When we program a recursion solution using formula (1), (2) and (3), the left-hand side of each formula is transformed to the signature of its implementing method while the right-hand side is transformed to the body of the method.
For example, (3) is transformed to pseudo-Python code
def f_star(n):
    return min(f_1(n) + x_1, f_2(n) + x_2)

So, when $f^\star$ is referenced, i.e., when f_star is called, $f_1[n]$ and $f_2[n]$ will be referenced, i.e., f_1(n) and f_2(n) will be called, where f_1(.) and f_2(.) will be explained below. Since we will call f_star(n) once, which is enough to get the value of $f^\star$, we get formula (4).
Formula (1) is transformed to pseudo-Python code
def f_1(j):
    if j == 1:
        return e[1] + a[1][1]
    else:
        return min(f_1(j - 1) + a[1][j], f_2(j - 1) + t[2][j - 1] + a[1][j])

So whenever $f_1[j]$ is referenced, i.e., when $f_1(j)$ is called, $f_1[j-1]$ will be referenced exactly one, i.e., f_1(j-1) will be called exactly once.
Similarly, whenever $f_2[j]$ is referenced, $f_1[j-1]$ will be referenced exactly once. (You can write the function f_2(j) explicitly yourself to check it out.)
Note that any reference to $f_1[j-1]$ must be brought by either a reference to $f_1[j]$ or a reference to $f_2[j]$. So we have
$$r_1(j-1) = r_1(j) + r_2(j).$$
Similarly or by symmetry, we also have
$$r_2(j-1) = r_1(j) + r_2(j).$$
Replacing $j$ by $j+1$, we get formula (5).