The 2 systems are equivalent. In one case you have a simple pendulum, in the other you have a compound pendulum. The fact that the pendulum string is flexible does not enter into the calculation. It could be replaced by a light rigid rod without affecting the outcome. The only difference then is the distribution of mass about the pivot point, which I think is the sole reason why conservation of linear momentum works for the simple pendulum but not the door.
In both cases you should use the conservation of angular momentum about the pivot point, not the conservation of linear momentum. Assuming that the force on the string or door from the pivot point acts towards the pivot (a central force) then there is no change in the angular momentum of the system during the instantaneous collision.
In both cases the initial angular momentum is $mvL$ where $L$ is the distance of the bullet from the pivot point. The final angular momentum is $J\omega_0$ where $J$ is the moment of inertia about the pivot and $\omega_0$ is the angular velocity immediately after the bullet is embedded, which is assumed to happen instantaneously.
For the simple pendulum, in which the mass $M$ of the pendulum is concentrated at distance $L$ from the axis, you have
where $m,M$ are the masses of the bullet and block and $v,V$ are the linear velocities of the bullet immediately before and of the bullet-and-block immediately after the collision. I have also used the fact that $V=\omega_0 L$. In this case the conservation of angular momentum is equivalent to the conservation of linear momentum.
For the compound pendulum, for which the mass is distributed like a rod up to a length $2L$ from the axis, you have
which is not the same as the result using the conservation of linear momentum in eqn (*) above.
[However, the conservation of linear momentum can be applied if the bullet strikes the door at the Centre of Percussion. (Thanks to Andrew Morton for pointing this out in his comment below.) The door swings on its hinge with the same period as a simple pendulum of the same mass concentrated at the CoP. The moment of inertia of the door can be written as $J=Mk^2$ where $k$ is the distance between the hinge and the CoP. So if the bullet strikes the door at the CoP then the conservation of angular momentum gives the same result as the conservation of linear momentum :
where now $V=\omega_0 k$ is the velocity of the CoP immediately after the collision.]
Both the simple and compound pendulums rotate about a fixed pivot. This motion can be decomposed into an instantaneous linear motion of the CM and a rotation about the CM. For the simple pendulum, the pivot is far outside of the bob, so the rotation of the bob about its CM is negligible compared with the motion of the CM. To a good approximation the impact results only in linear motion of the CM, so it is modelled well as a 1D linear collision. For the compound pendulum, the pivot point is not far from the CM compared with the size of the door, so the rotational motion about the CM is significant compared with the motion of the CM. The impact results in rotational as well as linear motion, so it cannot be approximated as a 1D linear collision.