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.
Your PV formula is right, but it applies to each amount paid so your second equation should be $$
5,000= \frac{1900}{(1+i)^1} + \frac{1900}{(1+i)^2} + \frac{1900}{(1+i)^3}
$$
Once you have calculated $i$ (by trial and error), then the outstanding balance after, say , 1 year just before the \$1900 repayment is $5000 (1 + i)$ so the outstanding balance just after the first repayment instalment is $$5000(1+i) - 1900.$$ You can repeat that process to get the outstanding balance after the second repayment instalment. You can check you answer by repeating it again and checking that the outstanding balance after the third repayment is zero.
See also related question repayment of loan with compound interest
Best Answer
As @saulspatz answered, the solution of $$k={1-v^{60}\over i}\qquad \text{with}\qquad v={1\over1+i}\qquad \text{and}\qquad k=\frac{ 100000}{1983.33 }$$ requires some numerical method.
However, you can make some approximation using Taylor series built at $i=0$; this would give $$k=60-1830 i+37820 i^2-595665 i^3+7624512 i^4-82598880 i^5+778789440 i^6+O\left(i^{7}\right)$$
Now, using series reversion, this would give $$i=x+\frac{62 }{3}x^2+\frac{9517 }{18}x^3+\frac{1979939 }{135}x^4+\frac{686499247 }{1620} x^5+\frac{5077158734 }{405}x^6+O\left(x^7\right)$$ where $x=\frac{60-k}{1830}$.
Using your value of $k$, this gives $x=\frac{189998}{36294939}\approx 0.00523483$ and a few terms of the above expansions will very quickly give $i\approx 0.00589$ which, multiplied by $12$, gives $0.07068$ as annual rate.
The exact solution, using Newton method, would be $i=0.0058900057$
Edit
For a shortcut evaluation of $i$, we could also use Padé approximants instead of Taylor series. To stay with simple equations to solve, let us use $$k=\frac{1-\frac{1}{(i+1)^{60}}}{i}\sim \frac{60-318 i+3422 i^2}{1+\frac{126 }{5}i+\frac{1953 }{10}i^2 }$$ and solving the quadratic equation in $i$ directly leads to $i\approx 0.00589003$.