Pullback of the Poincare bundle along $\varphi_{\mathcal L}$ is the Mumford bundle

Question

I'm trying to understand why $(1\times\varphi_{\mathcal L})^*(\mathcal P_A) = \Lambda(\mathcal L)$ where $\Lambda(\mathcal L) = m^*\mathcal L\otimes\operatorname{pr}_1^*\mathcal L^{-1}\otimes\operatorname{pr}_2^*\mathcal L^{-1}$ is the Mumford bundle. This is proved using the seesaw theorem; it's clearly true that $(1\times\varphi_{\mathcal L})^*(\mathcal P_A)|_{\{0\}\times A}$ is trivial, and similarly for the Mumford bundle. However, I'm not understanding why $(1\times\varphi_{\mathcal L})^*(\mathcal P_A)|_{A\times\{x\}}$ is $t_x^*\mathcal L\otimes\mathcal L^{-1}$. Why is this true?