As Daniel Fischer's Comment points out, your alternative requires a division with each iteration. Historically your alternative was the "standard" approach to applying Newton's method to finding square roots, being older by centuries than Newton's calculus! See for example the Babylonian method.
The slowness of division vs. multiplication on early mainframes motivated the division-free Newton iteration for solving $x^{-2} = a$, as noted in the Question:
$$ x_{n+1} = x_n(1.5 - 0.5 * a * x_n^2) $$
Despite the general acceleration of floating point operations on microprocessors, division continues to be significantly slower than multiplication on popular platforms. It follows that keeping the count of division operations low is still a valid design consideration.
Note that if $\sqrt{a}$ is really needed, rather than $1/\sqrt{a}$, we can get it with one additional multiplication:
$$ \sqrt{a} = a \cdot (1/\sqrt{a}) $$
Let's collect the comments into an answer. I like single letter variables for formulas, so
T -- $amount; or principal
n -- $numPay; number of months
R -- $payment; monthly rate
F -- $fee;
x -- $x; monthly interest rate
The payment plan, as I understand it, has an initial debt or principal $T$, and additional a fee $F$. After a month $R+F$ are paid, then every month $R$, after the n-th payment, the debt is paid in full.
As stated, the interest on the fee for the one month is not raised, which means that $F$ instead of $F\cdot(1+x_{nominal})$ is paid after a month. In balance, this still increases the effective interest rate.
The balance after $n$ months, seen from the start of the debt, and according to the flow of payments, reads as
\begin{align}T+F&=(R+F)(1+x)^{-1}+R(1+x)^{-2}+\dots+R(1+x)^{-n}\\
&=F(1+x)^{-1}+R\frac{1-(1+x)^{-n}}{x}\\
Tx+F\frac{x^2}{1+x}&=R(1-(1+x)^{-n})
\end{align}
Define
$$f(x)=Tx(1+x)+Fx^2-R(1+x-(1+x)^{1-n})$$
with derivative
$$f'(x)=T(1+2x)+2Fx-R(1-(n-1)(1+x)^{-n})$$
for a Newton iteration with the initial guess
$$x_0=\frac{nR-T}{\frac{n(n+1)}2R+F}$$
Best Answer
You have rediscovered the secant method.
The secant method a bit slower in the vicinity of the root than Newton-Raphson: its order is $1.618$ instead of $2$. However, since there is just one function evaluation per step (versus two for N-R: $f$ and $f'$), it may actually be faster. Depends on how complicated the derivative is.
This is much too broad to have an affirmative answer. Both methods converge from a vicinity of a root, if the function is reasonable. Both can fail to converge from further away. The basins of attraction of a root can be quite complicated (fractal), with tiny difference in initial position changing the outcome. Briefly: no, they are different methods, and you may find one method failing whether the other succeeds.