The intersection form on the nontrivial $S^2$-bundle over $S^2$, denoted $S^2 \tilde \times S^2$, can be written as $\left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$ with respect to a suitable basis for $H_2(S^2 \tilde \times S^2)$.$^\mathbf{1}$
Things get confusing when I try to use another basis. Let $[F]$ denote the homology class of the $S^2$ fiber and let $[B]$ come from a section $S^2 \to S^2 \tilde \times S^2$ of the bundle, which are linearly independent$^\mathbf{2}$. Since $[F] \cdot [B]=1$, they should form a basis for $H_2(S^2 \tilde \times S^2)$. It's clear that $[F]\cdot [F]=0$, so the intersection form is given by $\left[ \begin{smallmatrix} 0 & 1 \\ 1 & * \end{smallmatrix} \right]$, where $*=[B]\cdot [B]$. Now, regardless of what $[B]\cdot [B]$ is, there seems to be an issue: The conjugacy class of $\left[\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right]$ over $\mathbb{Z}$ contains matrices of the form $$\begin{bmatrix} 1 + 2ad & -2ab \\ 2cd & -1 -2ad\end{bmatrix},$$ and we can't have $0=1+2ad$ over $\mathbb{Z}$. So there can't be a change of basis connecting these forms.
What is wrong with the computation of the intersection form using the class of the fiber $[F]$ and base $[B]$?
Notes:
$S^2 \tilde \times S^2$ is diffeomorphic to $\mathbb{CP}^2 \# - \mathbb{CP}^2$. Each summand of $\mathbb{CP}^2 \# - \mathbb{CP}^2$ has its second homology generated by $\mathbb{CP}^1$, which has self-intersection $\pm 1$ in $\pm \mathbb{CP}^2$.
One "direct" way to see that sections exist is to use the homotopy lifting property, and one "indirect" way would be to use the fiber bundle LES in homotopy groups to see that $\pi_2(S^2 \tilde \times S^2) \to \pi_2(S^2)$ is surjective. For linear independence, we note that $p_*[F]=0$ while $p_*[B]$ generates $\pi_2(S^2)$.
The intersection matrix transforms like a quadratic form under change of basis, not like an endomorphism. That is, if $P$ denotes the change of basis matrix, the matrix $Q$ of the intersection form transforms by $Q \mapsto P^{t}QP$, not by the similarity $Q \mapsto P^{-1}QP$.