Regarding your main question, this is done in Cassels, Lectures on elliptic curves, $\S$ 8 (iv) p. 36. We may assume that the common rational point of the quadrics is $(X:Y:Z:T)=(0:0:0:1)$. Then the quadrics have the shape
\begin{align*}
Q_1 & = TL + R\\
Q_2 & = TM + S
\end{align*}
where $L,M$ (resp. $R,S$) are linear (resp. quadratic) in $X,Y,Z$. The polynomials $L,M$ are necessarily linearly independent, and eliminating $T$ gives the cubic equation
\begin{equation*}
P(X,Y,Z) = LS-RM=0.
\end{equation*}
So we are reduced to put a cubic into canonical form, which you know how to do.
Regarding your second question, it is indeed possible to work out the group law directly on the intersection $C$ of two quadric surfaces. This is explained in Husemöller, Elliptic curves, Introduction, $\S$ 8, p. 20. The idea is as follows: given two points $P,Q$ on $C$, consider the plane containing $O,P,Q$, and intersect it with $C$. By Bézout's theorem there will be 4 points of intersection, $O$, $P$, $Q$ and a fourth point $-(P+Q)$. The map $P \to -P$ is defined similarly.
If the two cubics $C_1=V(F_1)$ and $C_2=V(F_2)$ do not share a common component, then the ideal $I = (F_1,F_2)$ defines a $0$-dimensional subscheme $V(I)\subset \mathbb{P}^2$ length 9, which is a complete intersection of codimension 2.
In the fancy language above, the graded ring $\bigoplus_{n\geq0}I^n$ is generated as an algebra over $\mathcal{O}_{\mathbb{P}^2}$ by two elements $\xi_1,\xi_2$ in degree $1$, corresponding to the two equations $F_1,F_2$ generating $I$. They satisfy one relation $\xi_1F_2-\xi_2F_1=0$ corresponding to the syzyzy that holds between them.
Thus the blowup of $Z:=V(I)\subset \mathbb P^2$ is explicitly given by the projective variety
$$ \overline X = V(\xi_1F_2 - \xi_2F_1) \subset \mathbb P^1_{\xi_1,\xi_2}\times \mathbb P^2_{x,y,z}$$
and the morphism $\overline \pi\colon \overline X\to \mathbb{P}^1$ that you want to consider is simply the projection onto the first factor. (The blowup itself is given by projection onto the second factor.)
Unfortunately, if $Z$ is not reduced then this blowup $\overline X$ is going to be rather singular, so at this point you probably want to replace $\overline X$ with its minimal resolution $\mu\colon X\to \overline X$ and consider the induced morphism $\pi\colon X\to \mathbb P^1$.
I want to note that this construction is probably not exactly the approach Miranda had in mind. Judging from what he wrote he considers an iterative construction:
- first blow up the (reduced) intersection points of $C_1\cap C_2$
- look at the strict transform of $C_1$ and $C_2$ under this blowup and then blow up their intersection points,
- repeat until $C_1$ and $C_2$ are disjoint.
If you think hard enough about it, then I think it will follow from (a) the universal property of blowing up, and (b) the existence of minimal resolutions of surfaces, that this construction gives the same surface as $X$ above. The advantage of the first approach is it is obvious how to get the morphism to $\mathbb P^1$. The advantage of the second approach is that it gives a smooth surface directly, without having to go through some desingularisation process at the end.
Lastly, by some abstract nonsense there should exist some sheaf of ideals $\mathcal J$ on $\mathbb P^2$ such that the morphism $\pi\colon X\to \mathbb{P}^2$ is given by blowing up $\mathcal J$. Perhaps you had in mind to construct $\pi\colon X\to \mathbb{P}^2$ all in one go by working out how to write down this $\mathcal J$. I would think this is highly unlikely to be a pleasant calculation.
Best Answer
As explained in abx's comment, the canonical divisor of your surface is given by $$K_X=-3 \pi^*L + \sum E_i,$$ and this is precisely the class of $-\widetilde{C}_1$ (note that $C_1$ is a curve in $\mathbb{P}^2$, so I put the tilde to specify that we are considering its strict transform in $X$).
Now, $\widetilde{C}_1$ is the strict transform of a smooth cubic curve: since the self-intersection number of a cubic is $9$ and you are blowing up nine points on it, the self-intersection of the strict transform is $(C_1)^2=9-9=0,$ as claimed.
Finally, the strict transform of your pencil of cubics is a base-point free pencil of elliptic curves in $X$, providing an elliptic fibration $\pi \colon X \to \mathbb{P}^1$, endowed with nine sections corresponding to the nine exceptional divisors of the blow-up. By construction, the curve $\widetilde{C}_1$ is a fibre of $\pi$, hence it is a "vertical divisor", after all.