Consider a flat symplectic fibration $\pi: E \to B$, $B$ is complex 1-dim, no singular fiber, no curvature, but may have non-trivial monodromy. Suppose $(B, \omega_B = -d\lambda_B$ is a Liouville domain. pick a reference point $b_0 \in B$, and reference fiber $F = \pi^{-1}(b_0)$. Assume $F_0$ is also a Liouville domain, with $\lambda_F$, and symplectic parallel transport preserves the fiberwise Liouville 1-form structure near the boundary. A fibered Lagrangian $L_\pi$ is given by a base Lagrangian $L_B \In B$, assuming $b_0 \in L_B$, and a fiber Lagrangian $L_F \In F$, and parallel transport $L_F$ along $L_B$.
The question is, how do we make $L_\pi$ into a conical Lagrangian in the total space. First of all, it is not at all clear what does conical Lagrangian mean in the total space. We have to assume that, there is some relative 1-form $\lambda_\pi$, so that $\omega_\pi = d\lambda_\pi$ is a relative symplectic form for $\pi$. Then, we can do $\omega_{tot} = \omega_\pi + c_B \pi^* \omega_B$, where $c_B \gg 1$. This will not affect the symplectic connection, but helps make the horizontal distribution symplectic. So, we can consider the primitive $\lambda_{tot} = \lambda_\pi + c_B \lambda_B$. Suppose we already trivialized along the fiberwise boundary, we would have $\lambda_{tot}|_F = \lambda_\pi|_F$ and $\lambda_{tot}|_B = \lambda_\pi|_B + c_B \lambda_B$. The near the fiberwise boundary, the Liouville flow also factorize as a product $Z_F + Z_B$.
Near the fiberwise boundary, the fibered Lagrangian looks like $L_B \times L_F$, and $L_F$ is preserved (i.e. tangent to) $Z_F$. However $L_B$ is not tangent to $Z_B$, so we don't have a conical Lagrangian here. How to conicalize?
Maybe the question is: why do we want to conicalize? cannot we bound hol'c disk in some other ways? Like, fiberwise infinite wrapping of the Lagrangian? We hope so, but we don't know yet.
One way to control the boundary of the holomorphic disk to not escaping to infinity is to use a $J$-psh function $\varphi$, and require $L$ to be conical outside a compact subset, in the sense that $d^c \varphi$ restricted to $TL$ to vanish. The proof is that, take a large enough sublevel set of $\varphi$, so that outside this set, all Lag are conical, and there are no intersection points. Then, all hol'c disk has to be contained in that sublevel set. Because for a disk to escape to here, one either has to have an interior maxium or a boundary maximum, and both are forbidden.
To summarize, we can use J-psh function $\varphi$, conical Lagrangians (hence no intersections), to create a 'no-go' zone for holomorphic disks's $\varphi$-maximum. if we have a disk mapping to the defining domain of $\varphi$, then such a disk cannot touch the no-go zone. However, if a disk maps to a bigger space, only part of the disk is in the defining domain of $\varphi$, then the no-go result does not apply, since it can very well have no interior and boundary maximum of $\varphi$ on $D$.
If we have not just a fibration, but a product setup, then we do not need to make Lagrangian eventually sum-conical, we can use two psh functions to bound the two directions separately. But it requires us to maintain the product structure of $J$, $L$, and the auxiliary psh functions, in order to prove the compactness of moduli space. It shouldn't have to, the moduli space should remain compact under compactly supported perturbation of the interior data, like $J$ and $L$, just the sneaky proof breaks down. That's why we want to avoid using product structure to prove moduli space compactness.
why do we bother to attach the cylindrical end to the Liouville domain, and to extend the Lagrangian? if there is no intersections there, and no disks there, why not just forget them? because it was nice to have things complete to be safe, and then later you show you don't need those.
We start from a product space, $B^o \times F$, each factor an exact symp space, and we have product Lagrangians $L_B \times L_F$, disks in $L_B$ and $L_F$ can be bounded separately.
We consider another space $\pi: E \to B$, and $B^o \into B$, with trivialization $E^o :=\pi^{-1}(B^o) \cong B^o \times F$. We equip $E$ with a Kahler structure that when restricted to $E^o$ agrees with $B^o \times F$, namely $\varphi_E|_{E^o} = \varphi_{B^o} + \varphi_F$, also product type $J$.
We consider Lagrangians in $E^o$, of the same old product type. The 'only' change is that, we allow for much more disks in $E$, not just in $E^o$.
There exists psh $\varphi_B: B \to \R$ that extend $\varphi_{B^o}$.
It make sense to ask, whether $\pi_F^*\varphi_{F}: E^o \to \R$ can be extended to $E$ as a psh function. Sadly, the answer is no.