I was completely confused by that same explanation for a while. Don't even think of it like that. Think of it like this.
Definition: The effective rate of interest during the $n$th time period is
$$i_n = \frac{A(n) - A(n-1)}{A(n-1)}$$
Definition: The effective rate of discount during the $n$th time period is
$$d_n = \frac{A(n) - A(n-1)}{A(n)}$$
where $A(n)$ is the amount function (as defined in Kellison), the amount of money you have at time $n$. So, all you really need to understand here is that the rate of interest is a rate based on what you start with during the period. The rate of discount is a rate based on what you end up with. It's just two ways of looking at the same situation. There aren't a whole lot of real world situations where you borrow a bunch of money and immediately give some of it back. You would just borrow less.
By the way, that formula is all you need to calculate the effective rate of discount during period $n$ no matter what your $A(n)$ function is. So, in particular, it would work for your specific question of simple discount.
Question: Given a rate of 10% simple discount, calculate the effective rate of discount during period 5.
Answer: If we have 10% simple discount, then we know our accumulation function is $a(t) = \frac{1}{1 - 0.1t}$ for $0 \leq t < \frac{1}{d} = 10$. This is basically the definition of simple discount. If you have simple discount, this is your accumulation function. Memorize that. Then use it.
Therefore
$$d_5 = \frac{a(5) - a(4)}{a(5)} = \frac{2-10/6}{2} = \frac{1}{6} = 16.666666... \%$$
If you wanted to calculate the effective rate of interest when you are given the effective rate of simple discount, you can do that too. For example, in this same example,
$$i_5 = \frac{a(5) - a(4)}{a(4)} = \frac{2-10/6}{10/6} = \frac{1}{5} = 20 \%$$
Nothing changed. We're just looking at the same problem differently. In the discount case, how much money did we earn that period relative to how much we had at the end? In the interest case, how much money did we earn that period relative to how much we had at the beginning.
Outline: The PV in the first option is $6000+\frac{5940}{1+i}$. After all, the PV of the first payment of $6000$ is exactly $6000$, since it is being paid now.
The equation you end up with, after multiplying through by $1+i$, is a quadratic in $\sqrt{1+i}$. Let $x=\sqrt{1+i}$ and solve.
Best Answer
Let d be the discount rate, t be the total time of investment, and k be the largest integer such that t > k . First we calculate the compound interest over time k, then multiply that by the simple interest for time t-k. Thus your accumulation function is:
$ a(t) = (1 -d)^{-k} (1-d(t-k)) $
Similarly, the accumulation function for compound interest with interest rate i :
$ a(t) = (1 + i)^k (1 + it) $
Bonus points if you don't forget to take $ (a(t))^{-1} $ for present value in your question like I did.