Is there more to this relationship with the Fibonacci numbers?

You have builе a Fibonacci (?) field $Q[x]/(x^2-x-1)$. Every element of this field can be written as binomial, so it has representation in 2x2 matrices: $$ ax+b\qquad\Leftrightarrow\quad \begin{pmatrix}a+b & a\\ a & b\end{pmatrix} = aF+bI $$ You can check that $F^2=\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}^2=F+I $

You $O(\log n)$ algorithm is a well known the Doubling Method, which is essentially fast multiplication algorithm for matrix $F$.

To generalize, you should just select right first element. So in ordinary fibonacci, you start with (1,0), which is $X_1=1x+0$, next element is $X_2=x X_1$, then $X_3=x^2 X_1$ and so on. If you want to start with (-1, 2), then $Y_1=-1x+2$, $Y_2=x Y_1$, $y_3=x^2 Y_1$ and so on. You can use the same algorithm of fast multiplication to find $Y_n$.

Edit. As an example, I want to show how to build a tribonacci field $Q[x]/(x^3-x^2-x-1)$.

If we have $y=\alpha x^2+\beta x+\gamma$, then $$xy = \alpha x^3+\beta x^2+\gamma x = (\beta+\alpha)x^2 + (\gamma+\alpha)x+\alpha$$ So element $x$ is equivalent to matrix: $$ T = \begin{pmatrix}1 & 1& 1\\ 1 & 0&0\\0&1&0\end{pmatrix}. $$ So our field can be represented with $Q[I, T]$