Doubts about IIR filters definition and transfer function
No. In this case, "infinite" really means semi-infinite &mdash; nonzero for positive time only. If it were nonzero for any negative time, then it would be non-causal, and therefore not physically realizable.
Your equation doesn't look correct &mdash; you have \$y(n-k)\$ on both sides, but I think you intended to have \$x(n - k)\$ on the right. Also, it obscures the term for the actual output (\$y(n)\$). It would normally be written more like this:

$$y(n) = \sum_{k=0}^{M-1}b_k x(n-k) - \sum_{k=1}^{L-1}a_k y(n-k)$$

It's just one way of showing the relationship between the "feedforward" terms (\$b_k\$) and the "feedback" terms (\$a_k\$). Combining all of the y terms together in the form that you showed makes it easier to take the z transform, which then allows you to characterize the filter as

$$H(z) = \frac{B(z)}{A(z)}$$
See above. The actual output term is just \$y(n)\$ (where k=0 and a0 is normalized to 1). The rest (k=1..L-1) represent delayed (older) output values being used for feedback.




Note that n = 0 ... N represents a sequence of input or output samples, with time increasing with N (toward the future). Therefore, the expression (n - k), with k positive, represents looking toward the past, to older samples of the input and output.