Does the order of the Fibonacci sequence's initial values matter?
In the particular case of the Fibonacci number sequence OEIS A000045 (or series) there is some difference of opinion as amply evidenced by the Wikipedia article and OEIS entry. The most common convention is that $\,F_0=0\,$ and $\,F_1=1\,$ and this choice has much in its favor. This choice implies that its generating function is $$ \sum_{n=0}^\infty F_n\,x^n = \frac{x}{1-x-x^2}. $$ Notice the $\,x\,$ in the numerator. For this and a few other reasons, some prefer the choice $\,F_0=F_1=1\,$ instead, which implies that the generating function is now $$ \sum_{n=0}^\infty F_n\,x^n = \frac{1}{1-x-x^2}. $$ Both choices are valid and useful in some cases and not as useful in other cases. Thus, it is a good idea to be clear on which choice is being used in any given context since the choice will usually matter.
The situation with negative index Fibonacci sequence elements is that the recurrence relation for the sequence can be used to uniquely extend the sequence in the negative index direction. For the common convention this implies that $$ F_{-n} = (-1)^{n-1}F_n \quad\text{ for all integer }n. $$ The result for the other convention it is that $$ F_{-n} = (-1)^n F_{n-2} \quad\text{ for all integer }n. $$
What if we choose arbitrary starting values with the same recurrence relation? That is, if we choose $\,F_0=a\,$ and $\,F_1=b,\,$ what do we get? The recurrence relation is linear and this implies that the sequence of numbers we get is a linear combination of the two initial value choice sequences. The generating function now is $$ (a) \!+\! (b)x \!+\! (a+b)x^2 \!+\! (a+2b)x^3 \!+\! \cdots \!=\! \frac{a\!+\!(b\!-\!a)x}{1-x-x^2}. $$
Note that initial values of Fibonacci and related sequences can be given for any two consecutive index values--not just for $\,a_0\,$ and $\,a_1\,$ since the recursion going forward and backward can reach any other integer index value. Again, this is just a matter of convenience and convention. In the case you cite, specifying $\,a_{-1}\,$ and $\,a_0\,$ is fine. The initial values $\,a_{-1}=1,\,a_0=0\,$ gives the same sequence as $\,a_0=0,\,a_1=1.$ This special case may be misleading though. The initial values $\,a_{-1}=u,\,a_0=v\,$ gives the same sequence as $\,a_0=v,\,a_1=u+v\,$ and $\,u+v\,$ is equal to $\,u\,$ if and only if $\,v=0.$
The recursion $a_n=a_{n-1}+a_{n-2}$ is "encoded" in the denominator of the generating function: $1-x-x^2$. The numerator is determined by the values of $a_0$ and $a_1$. In particular, $$ \overbrace{\left(1-x-x^2\right)\vphantom{\sum^\infty}}^\text{denominator}\overbrace{\sum_{n=0}^\infty a_kx^k}^{\substack{\text{generating}\\\text{function}}}=\overbrace{a_0+(a_1-a_0)\,x\vphantom{\sum^\infty}}^\text{numerator}+\sum_{n=2}^\infty\overbrace{(a_n-a_{n-1}-a_{n-2})\vphantom{\sum^\infty}}^0\,x^n $$ Thus, the generating function is $$ \frac{a_0+(a_1-a_0)x}{1-x-x^2} $$
I prefer the definition of the Fibonacci Numbers that starts with $F_0=0$ and $F_1=1$, whose generating function is $$ \frac{x}{1-x-x^2} $$ because, among other things, we have $$ n\mid m\implies F_n\mid F_m $$ Furthermore, $$ F_{-n}=(-1)^{n-1}F_n $$ Lucas Numbers also obey the same recursion as Fibonacci Numbers, and their generating function is $$ \frac{2-x}{1-x-x^2} $$
There is already a nice description of generating functions in Somos answer. I will focus here on the question you had about the author's general formula.
$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$
You are right in saying, 'for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$.' You are also correct in observing that you weren't given $a_{-2}$ and the formula requires it.
Also, technically, since the recursion given in the exercise only holds for $n > 0$, $a_{-2}$ is indeterminate in the exercise. So this is 'bad' notation on the author's part if you are expected to use this general theorem, in my opinion.
So, to alleviate this I propose you consider an extension to the recursion, where the recursion holds for $n\geq 0$, (as it must whenever the theorem you have stated and it's associated formula should apply). In such a case, although you aren't given $a_{-2}$ you can find its value from the recursion itself since:
$$a_{0} = a_{-1} + a_{-2}$$
So $a_{-2} = a_{0} - a_{-1} = -1$. Plugging it all in:
$$ G(x) = \frac{x(x^{-1}) + x^{2}(-x^{-2}+ x^{-1})}{1-x-x^{2}} = \frac{x}{1-x-x^{2}} $$
So, it yields the expected generating function for the case where $a_{0}=0$ and $a_{1} = 1$.
So, I understand the confusion you had. However to answer the question as to what happened, when you shift the indices for the initial condition, it will result in shifting the indices throughout the sequence which emerges from the recursion and this will change the generating function.