Let $\mathcal{X}$ be an elliptic surface over $\mathbb{P}^1$ without a section and let $\mathcal{J}$ be an elliptic surface over $\mathbb{P}^1$ with a section. Assume we have the commutative diagram
\begin{array}{ccc} \mathcal{X} \times_{\mathbb{P}^1} C & \hspace{-0.2cm} \stackrel{\Large{\varphi}}{\leftarrow - \rightarrow} \hspace{-0.2cm} & \mathcal{J} \times_{\mathbb{P}^1} C \\ pr_{\mathcal{X}} \downarrow & \Box & \downarrow pr_{\mathcal{J}} \\ \mathcal{X} \longrightarrow \hspace{-1.6cm} & \mathbb{P}^1 & \hspace{-1.7cm} \longleftarrow \mathcal{J} \end{array}
In this diagram $C$ is a curve over $\mathbb{P}^1$, $\varphi$ is a birational isomorphism over $C$, and $\times_{\mathbb{P}^1}$ is the fiber product over $\mathbb{P}^1$.
Is $\mathcal{J}$ the Jacobian of $\mathcal{X}$ in that case?
The question arised from the following example. Let $\mathcal{X} = Q_1 \cap Q_2 \subset \mathbb{A}^4$ over an algebraically closed field $k$ of even characteristic, where
\begin{array}{l} Q_1\!: y^2 + y = t[(x^{\prime^2} + xx^\prime t + x^\prime t^2 + xt^3 + t^4) + (x^\prime + xt + t^2)], \\ Q_2\!: x^\prime = x^2. \end{array}
The projection on $t$ is the elliptic fibration, because a general fiber is an nonsingular intersection of two quadrics.
I know that $\mathcal{X}$ is rational. I can write the proof if necessary.
Consider the conic $C\!: s^2 = t \subset \mathbb{A}^2$. You can check that the surface $\mathcal{X} \times_{\mathbb{P}^1} C$ is birationally isomophic over $C$ to the surface $\mathcal{J}\times_{\mathbb{P}^1}C$ by the rational map $$ \varphi: \mathcal{J}\times_{\mathbb{P}^1} C \to \mathcal{X} \times_{\mathbb{P}^1} C, \qquad x := \frac{x_2 + s}{t^2}, \ y := \frac{y_2}{t^2} + (t^2 + 1 + x^2)s, \ x^\prime := x^2, $$ where $$ \mathcal{J}\!: y_2^2 + t^2y_2 = x_2^3 + t(t^2 + 1)x_2^2 + t(t^5 + t^3 + 1)x_2 + t^2(t^5 + t^3 + t^2 + 1). $$ It is clear that Kodaira-Néron model of this surface is the K3 surface. However in the book Enriques surfaces I by F. R. Cossec and I. V. Dolgachev the proposition 5.6.1.(ii) states that the Jacobian of an rational elliptic surface is again rational. Thus I have the contradiction, because a K3 surface is irrational. What is an error?
