The two blow-ups and the quartic pencil

Proof. Choose quadrics $F(X,Y,Z)$ and $G(X,Y,Z)$ restricting to $f$ and $g$ on $Q$, and write $q=XZ-Y^2$. Then $C_U$ is the complete intersection $$q=uF+vG=0.$$ Since $C_U$ is the graph of $\eta_U$, it is smooth. Thus $X_U$ is smooth. The quadrics $q,F,G$ have no common zero: a common zero would belong to $Q$ and be a common zero of $f$ and $g$.

In $\mathbb P^2_{a,b,c}\times\mathbb P^2_{X,Y,Z}$, put $$Y_U=\{aq+bF+cG=0\},\qquad O=[1:0:0].$$ The second projection of $Y_U$ is the projectivization of a rank-two vector bundle, since $q,F,G$ have no common zero. In particular, $Y_U$ is smooth. The fibre of its first projection over $O$ is $Q$. Let $\mathbb F_1=\operatorname{Bl}_O(\mathbb P^2)$, with ruling $\mathbb F_1\to\mathbb P^1_{u,v}$ given by $[b:c]$. The proper transform of $Y_U$ in $\mathbb F_1\times\mathbb P^2$ is $X_U$. Indeed, writing $b=\lambda u$ and $c=\lambda v$, its equation is $$aq+\lambda(uF+vG)=0.$$ Over $Z$ this is the usual incidence equation for the blow-up of the ideal $(q,uF+vG)$. Over $Y_U$ it is the blow-up of its smooth fibre over $O$. Its exceptional divisor $E$ is the strict transform of $\mathbb P^1\times Q$, hence $E\cong\mathbb P^1\times\mathbb P^1$.

Let $H_1$ be the pullback of a line from $\mathbb P^2_{a,b,c}$, let $H_2$ be the pullback of a line from $\mathbb P^2_{X,Y,Z}$, and let $H_3$ be the pullback of a point from $\mathbb P^1_{u,v}$. Adjunction in $\mathbb F_1\times\mathbb P^2$ gives $$-K_{X_U}=H_1+H_2+H_3.$$ The divisor $H_1+H_3$ is ample on $\mathbb F_1$, and $H_2$ is ample on the second factor, so their sum is ample. On $\mathbb F_1$ we have $H_1^2=H_1H_3=1$ and $H_3^2=0$. Since $[X_U]=H_1+2H_2$, we obtain $$(H_1+H_2+H_3)^3(H_1+2H_2) =6(H_1+H_3)^2+3H_1(H_1+H_3)=24.$$ Finally, the two equations defining $C_U$ have bidegrees $(0,2)$ and $(1,2)$. Their restrictions to $C_U\cong\mathbb P^1$ have degrees $4$ and $8$. Thus $$N_{C_U/Z}\cong\mathcal O_{\mathbb P^1}(4)\oplus \mathcal O_{\mathbb P^1}(8),$$ so the exceptional divisor $E'$ of $\pi$ is $\mathbb F_4$. ◻

Proof. The backward implication follows by extending an automorphism of $Q$ to $\mathbb P^2$ and changing the basis of the pencil. For the converse, $\rho(X_U)=3$, and the nef cone is generated by $H_1,H_2,H_3$. To see the latter assertion, these divisors are nef, and the fibres of $\pi$, the fibres of $\sigma$, and the fibres of $X_U\to\mathbb F_1$ have intersection vectors proportional to $(1,0,0)$, $(0,0,1)$, and $(0,1,0)$, respectively. Consequently these are the three extremal rays of the cone of curves. The first two contractions are divisorial and the third is a conic bundle. Their exceptional divisors $\mathbb F_4$ and $\mathbb P^1\times\mathbb P^1$ are not isomorphic. An isomorphism therefore preserves the contraction $\pi$ and descends to an isomorphism of the pairs $(Z,C_U)$ and $(Z,C_{U'})$. An automorphism of $Z$ preserves its two projections and must preserve the conic which is the image of the centre in $\mathbb P^2$. Its action on this conic belongs to $\operatorname{PGL}_2$; its action on the first factor merely changes the basis of the pencil. ◻

Proof. Near a point of $C_U$, choose a coordinate $z$ on $Q$, a normal coordinate $w$ to $Q$, and a coordinate $t$ on the first factor. The centre has ideal $(w,t-h(z))$. At a ramification point of index $e$, analytic coordinates give $h(z)=z^e$. In one blow-up chart the equation is $t=z^e+\lambda w$. The fibre $t=0$ thus has equation $\lambda w+z^e=0$. It is smooth for $e=1$ and has type $A_{e-1}$ otherwise. The other chart has coordinates $(t,z,\mu)$ and equation $w=\mu(t-z^e)$, so its fibres are smooth. The strict transform of $\mathbb P^1\times Q$ lies in the latter chart and is given by $\mu=0$. Thus it contains none of these singular points. ◻