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The question about secant lines reduces to a Chern class computation on the blowing up $\nu:\widetilde{\Sigma \times \Sigma}\to \Sigma \times \Sigma$ along the diagonal, where $\Sigma$ is the cubic scroll. Your claim is equivalent to the assertion that in $A^4(\widetilde{\Sigma \times \Sigma})$, the following class equals $2$ (overcounting by $2$, because the pair of points is ordered). $$2 = a^4 - 3a^2 b + b^2,\ \ a = (\nu^*\text{pr}_1^*c_1(\mathcal{O}(1)) + \nu^*\text{pr}_2^*c_1(\mathcal{O}(1)) - e),$$ $$b = \nu^*(\text{pr}_1^*c_1(\mathcal{O}(1))\cdot \text{pr}_2^*c_1(\mathcal{O}(1)) - i_*\pi^*c_1(\mathcal{O}(1)) $$ where $i:E\to \widetilde{\Sigma\times \Sigma}$ is the exceptional divisor, where $\pi:E\to \Sigma$ is the projection to the diagonal, and where $\mathcal{O}(1)$ is the usual Serre twisting sheaf on $\mathbb{P}^5$. I have not computed whether or not $a^4-3a^2b+b^2$ does indeed equal $2$ . . .