Tacitely, I am working over the field of complex numbers!
Let $\overline{M}_{0,4}\cong\mathbb{P}^1$ be the compactification of the moduli space of $4$-pointed stable rational curves. The relevant universal family $(\overline{U}_{0,4},\widetilde{\tau}_1,\widetilde{\tau}_2,\widetilde{\tau}_3,\widetilde{\tau}_4)$ is the blow-up of $\mathbb{P}^1\times\mathbb{P}^1$ at three points $\{(0,0),(1,1),(\infty,\infty)\}$.
For clarity $pr_1:U_{0,4}=\mathbb{P}^1\times\mathbb{P}^1\to\overline{M}_{0,4}$ has $4$ canonical sections: \begin{equation} \tau_1(x)=(x,0),\tau_2(x)=(x,1),\tau_3(x)=(x,\infty),\tau_4(x)=(x,x) \end{equation} wich intersect at the previous three points; in other words, these are not disjoint section.
By all this: $\widetilde{\tau}_i$ is the strict transformation of $\tau_i$.
For me it is ok that these new sections are disjoints, but I have no geometric intuition:
(Q1) can someone give it one?
Another construction of $\overline{U}_{0,4}$ is the following: considering the fundametal points $\{P_0=[1:0:0],P_1=[0:1:0],P_2=[0:0:1],P_3=[1:1:1])$ of $\mathbb{P}^2$ and the pencil of conics $sX(Y-Z)+tZ(X-Y)=0$ through these points, with $[s:t]\in\mathbb{P}^1$.
The fundamental points of $\mathbb{P}^1$ determine the all and only singular conics of the pencils. The blow-up of $\mathbb{P}^2$ at these base points of the pencil is isomorphic to $S_5$, the degree $5$ Del Pezzo surface. This construction resolves the rational map \begin{equation} \varphi:[x:y:z]\in\mathbb{P}^2\dashrightarrow[z(x-y):x(z-y)]\in\mathbb{P}^1 \end{equation} defined by the previous linear system.
The unsolved point is the following:
For each each base point, show that there is exactly one conic in the pencil with a given tangent direction, and conclude that the four exception divisors provide four disjoint sections of the resolved map $\widetilde{\varphi}:S_5\to\mathbb{P}^1$.
In other words:
(Q2) for a base point (e.g. $P_1$) there is exactly one conic in the pencil with a given tangent direction, but I can not exclude that for another base point (e.g. $P_2$) there is another conic in the pencil with the same tangent direction: is it correct?
(Q3) How can I define four (disjoint) sections of $\widetilde{\varphi}$ following this construction?
I understand by this topic that $S_5\cong\overline{U}_{0,4}$ as projective varieties, but I do not undestand how can I contruct the previous sections?
Reference: J. Kock, I. Vainsencher (2007) An Invitation to Quantum Cohomology, Birkhäuser; exercise 2.5.(iv)
Q1: first, lets pass to $\mathbb A^1 \times \mathbb A^1 \subset \mathbb P^1 \times \mathbb P^1$ with coordinates $(x,y)$. We want to blow up the origin and show that this makes $\tau_1 = V(y)$ and $\tau_4 = V(x-y)$ disjoint.
For one chart of the blowup, let $x = u$ and $y = uv$. Then the total transform of $\tau_1$ is defined by $uv=0$ and for $\tau_4$ by $u - uv = u(1-v)$. The exceptional divisor is $u=0$, $\tilde \tau_1$ is $v=0$, and $\tilde \tau_4$ is $v=1$ hence clearly disjoint. The calculation in the other chart is almost exactly the same, and the calculations for the other two blown up points are the same after changing coordinates to bring $(1,1)$ and $(\infty,\infty)$ to the origin.
Q2: two conics can indeed have the same tangent direction at different base points, but this does not actually matter; the "same tangent direction at a different point" is not the "same direction" in any sense that will be "noticed" by blowing up the base points.
Q3: the exceptional divisors furnish the sections in this construction (hence the equations of the exceptional divisors are the equations of the sections in a given coordinate system). Since they lie over disjoint points, they are themselves disjoint, and since a given tangent direction at a given base point is only realized by a single conic (i.e. fiber of the family), that conic will be the only fiber to meet the section at the point corresponding to that tangent direction (so in other words, these are genuinely sections of a fiber bundle).