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
$$x_0=\frac{nR-T}{\frac{n(n+1)}2R+F}$$
The general idea of Newton-like methods is that you approximate the equation by some linearization. This linearization does not need to be very exact to still get a reasonably fast convergence. Here just iterating the stage equations as a fixed-point iteration gives linear convergence with a factor $O(h)$. If the linearization is also $O(h)$ exact, then the resulting Newton-like iteration will still be linear, but now with a factor $O(h^2)$. Depending on how the initial guess, the predictor, is computed, this can result in a drastic reduction of required corrector steps in the Newton-like iteration to get to an error level that is below the method step truncation error.
$\newcommand{\D}{\mathit{\Delta}}$
So if $J$ is the Jacobian at $y_n$, then you can decompose $f(y+\D y)=f(y)+J\D y+R(\D y)$. For the given 2-stage method this gives the system
$$
\vec k_1-hJ(B_{11}\vec k_1+B_{12}\vec k_2)=f(\vec y_n)+R(h(B_{11}\vec k_1+B_{12}\vec k_2))
\\
\vec k_2-hJ(B_{21}\vec k_1+B_{22}\vec k_2)=f(\vec y_n)+R(h(B_{21}\vec k_1+B_{22}\vec k_2))
$$
In an iteration step the left side is a linear system for the new values, while the right side is computed from the old values. This form demonstrates that the right side is constants plus small superlinear corrections, in the practical computation the right side will be computed as $f(y)+R(\D y^{old})=f(y+\D y^{old})-J\D y^{old}$.
\begin{align}
\pmatrix{I_2-hB_{11}J&-hB_{12}J\\-hB_{21}J&I_2-hB_{22}J}
\pmatrix{\vec k_1\\\vec k_2}
=
\pmatrix{f(y+\D_1 y^{old})-J\D_1 y^{old}\\f(y+\D_2 y^{old})-J\D_2 y^{old}}
\end{align}
Using a Kronecker product, the matrix on the left could also be written as $I_4-hB\otimes J$.
Best Answer
Put:
$$X = \log\left(\frac{\alpha}{\alpha + \beta}\right) = \log\left(\frac{1}{1 + \frac{\beta}{\alpha}}\right)$$
Then working in terms of X will ensure that the logarithm is always well defined. You then need to define another independent variable Y such that a linarization of Y in terms of small changes in $\alpha$ and $\beta$ does not become almost linearly dependent on the way the change in X depends on small changes in $\alpha$ and $\beta$. Since X depends on the ratio of $\alpha$ and $\beta$, you can choose Y to be a function of the product of $\alpha$ and $\beta$, so:
$$Y = \alpha\beta$$
might work well, at least you'll have elminated two potential problems with Newton-Raphson.