Fibonacci number vector space

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If we define a vector space of sequences over real field which satisfy the Fibonacci number condition i.e. $f(n) = f(n-1) + f(n-2)$ with any $n \in \mathbb{Z}_+$. Now the book says that, since every sequence is uniquely determined by the two terms $f(0)$ and $f(1)$ the dimension of the space is 2. I do not get this part completely. The members of the vectors space is the entire sequence. The sequence $(0,1,...), (1,0,....), (2,3,...)$ are all different sequences.

How can just the first two elements uniquely define the sequence ? I can see that after some point all the sequence have the same subsequence. But that does not mean that sequences are same, right ? How can I show that the sequence $a = (1,0,1,1,2,3,...)$ with $f(0) = 1$ and $f(1) = 0$ and $b = (0,1,1,2,3,...)$ with $f(0) = 0$ and $f(1) = 1$ are basis for this space ?

Let us say there is a sequence $c = (2,3,5,8,13,...)$ with $f(0) = 2$ and $f(1) = 3$, how can I write c as a linear combination of a and b ?

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If every sequence is uniquely determined by f(0) and f(1), you just need to focus on the first two terms of each of the sequences in play here. How can you write the vector (2,3) as a linear combination of (1,0) and (0,1)?