Difference Equation - Time Response analysis of Discrete Systems
The answer to your first question (why $j$ starts at $1$) is contained in your first sentence: the output is computed from the current input value ($i=0$), from past input values ($i\ge 1$), and from past output values (i.e. $j\ge 1$). Imagine that you had the following difference equation with $j$ starting at $0$:

$$y(n)=\sum_{i=0}^Mb_ix(n-i)-\sum_{j=0}^Na_jy(n-j)\tag{1}$$

Then you could rewrite (1) as

$$y(n)=\sum_{i=0}^Mb_ix(n-i)-\sum_{j=1}^Na_jy(n-j)-a_0y(n)\tag{2}$$

and consequently

$$y(n)(1+a_0)=\sum_{i=0}^Mb_ix(n-i)-\sum_{j=1}^Na_jy(n-j)\tag{3}$$

Now you can divide (3) by $1+a_0$ to arrive at an equation like the one given in your question (with $j$ starting at $1$) (and with all coefficients scaled by the constant $1+a_0$). So you can always use the difference equation with $j$ starting at $1$ without loss of generality.

Concerning the 3 examples of difference equations you gave I'll just discuss the first one, the others are totally analogous. Here you have $M=5$ and $N=1$ with coefficients $b_1=0.002$, $b_5=0.01$, and $b_0=b_2=b_3=b_4=0$, and $a_1=-0.99$.

Finally, concerning your last question about the system $y(n)=2x(n)$. That's again a special case of your general difference equation with all $a_i=0$ (i.e. no recursion) and with $b_0=1$ (and all other $b_i=0$, $i\ge 1$). So all your examples are special cases of the general difference equation.