For your first question, maybe this can help (esp. section 7): https://arxiv.org/abs/1307.1771
An intuitive idea goes as follows: if $C_1: g(x:y:z)=0$ and $C_2: h(x:y:z)=0$ are two of the cubics passing through the 9 given points, then for all scalars $\alpha,\beta$, the cubic of equation $\alpha.g(x:y:z) + \beta.h(x:y:z) = 0$ also goes through the 9 points. In fact, one can prove that all the cubics through these 9 points have such an equation and, thus, each of these cubics correspond to a point $[\alpha : \beta]\in\Bbb P^1$.
One can prove that these cubics cover all the projective plane so this should give a map $\Bbb P^2\to \Bbb P^1$ which maps a point to the $[\alpha:\beta]$ corresponding to the cubic containing this point. But, what value of $[\alpha:\beta]$ can we choose if this point is one of the intersection points of the 9 cubics (and hence of all our cubics)? We can't know! That's why we obtain only a rational map $\Bbb P^2\dashrightarrow\Bbb P^1$.
Now, blowing up these 9 points removes indeterminacies of this map, since in the surface $S$ obtained with this blow-up, each of these points is replaced by a $\Bbb P^1$. Pulling back all our cubics of $\Bbb P^2$ on $S$ gives us a family of elliptic curves which don't intersect anymore and we obtain an elliptic fibration $S\to \Bbb P^1$, i.e. a family of elliptic curves indexed by $\Bbb P^1$ with some singular fibers (like nodal or cuspidal cubics). But the smooth fibers are elliptic curves, thus one can obtain translations on them because of their group law.
The $\Bbb Z^8$ comes from this: the 8 sections can give translations on each smooth fiber (simply between the points of intersection between these sections and the fiber) and if our 9 points aren't in special positions, then these translations (which only give birational self-maps of $S$ in general, because translations aren't defined on singular fibers) can be extended on the whole surface $S$ to give automorphisms of $S$ which are "sufficiently independent" to give a subgroup of $\text{Aut }S$ isomorphic to $\Bbb Z^8$.
Of course, many points are more or less fuzzy in my story (and to be honest, I don't remember how to prove all the details), but the paper I mentioned above explains many things about these surfaces and gives precise statements and proofs, but needs some background on algebraic surfaces like divisors and Riemann-Roch theorem.
For your second question, see https://mathoverflow.net/questions/8812/why-do-automorphism-groups-of-algebraic-varieties-have-natural-algebraic-group-s