The reason the end correction impedance is imaginary (not real) is because it represents energy dissipation from the pipe, by radiation into the infinite region of fluid (air) outside the pipe.
Since it dissipates energy, you can think of it physically as a type of damping. All types of damping is represented by imaginary terms in the transfer function - see any textbook or web site on damped harmonic motion for examples.
The impedance is a property of the medium, not of "the wave". The relevant properties include the geometry of the region - i.e. the diameter of the pipe, but not the properties of the wave itself. For example the end correction is independent of the wavelength of sound waves in the pipe.
In a first course (high school level) on sound waves in pipes, you were probably told that at the "open end" of a pipe the pressure is zero and the velocity is non-zero. In fact that is only approximately correct. If it was exactly true, there would be no work done on the air outside the pipe, and you would not hear any sound produced by the pipe. (Work = $\int P\, dV$, which is $0$ if $P = 0$).
Levine and Schwinger produced a series approximation which is a better approximation to the true end condition, but it is still not exact in real world situations. For example, it ignores the viscosity of the air, which creates a boundary layer in the flow inside the pipe and therefore means that the axial flow velocity over a cross-section through the pipe is not constant.
If the wavelength of the wave in the pipe short, with the same order of magnitude as L&S's end correction formula, the approximation breaks down and the L&S formula becomes "completely wrong," but that situation only arises for a very short pipe where the length is similar to the diameter, or for a high harmonic of the oscillation in a longer pipe.
Physically, you can think of the end correction as representing a cylindrical "lump" of air outside the end of the pipe (with length = the end correction, diameter = the pipe diameter) which is forced to oscillate by the vibrating air inside the pipe, and then radiates all its sound energy away into free space.
Note, the math in the L&S derivation is quite "advanced", and their overall approach to solving the problem predates the use of numerical methods (in 1948, computers had recently been invented, but had not yet become commonplace or powerful enough to attack problems like this) - so don't worry about the fact that you don't understand it, unless you are currently studying "theoretical physics" at graduate level! If you set up a computer model to calculate the 3-D flow pattern in and around the pipe using current methods (e.g. the Boundary Element method for acoustic problems) you will get the "end correction" included automatically because of the "no energy reflection" boundary conditions you specify for the flow field infinitely far from the pipe, without having to do anything "special" to include it in the model.
You might find this paper interesting and perhaps more accessible than the original L&S paper: https://arxiv.org/pdf/0811.3625.pdf
Best Answer
First, I believe you have a mistake in your equations. The transmission should be $$T = \frac{2Z_0}{Z_1 + Z_0}.$$
The transmission in Scenario 1 is the transmission through the one boundary:
$$T_1 = \frac{2Z_W}{Z_G + Z_W},$$
while the total transmission in Scenario 2 is the product of two boundaries:
$$T_2 = \frac{2Z_W}{Z_P + Z_W} \frac{2Z_P}{Z_G + Z_P}.$$
Using the corrected equations, the transmissions are 11.8% in Scenario 1 and 14.6% in Scenario 2.
I was curious if Scenario 2 could ever give a lower transmission, so I worked it out. If we assume $T_1$ is bigger, we can figure out under what circumstances that is satisfied as follows:
$$T_1 \stackrel{?}{>} T_2,$$
$$\frac{2Z_W}{Z_G + Z_W} \stackrel{?}{>} \frac{2Z_W}{Z_P + Z_W} \frac{2Z_P}{Z_G + Z_P},$$
$$Z_W(Z_P+Z_W)(Z_G+Z_P) \stackrel{?}{>} 2 Z_W Z_P(Z_G+Z_W),$$
$$(Z_P+Z_W)(Z_G+Z_P) \stackrel{?}{>} 2 Z_P (Z_G+Z_W),$$
$$Z_P Z_G + Z_W Z_G + Z_W Z_P + Z_P^2 \stackrel{?}{>} 2 Z_P Z_G + 2 Z_P Z_W,$$
$$ Z_W Z_G + Z_P^2 \stackrel{?}{>} Z_P Z_G + Z_PZ_W,$$
$$ Z_P^2 - Z_P (Z_G + Z_W) + Z_W Z_G \stackrel{?}{>} 0,$$
$$ (Z_P - Z_G)(Z_P - Z_W) \stackrel{?}{>} 0.$$
By inspection we see that $Z_P = Z_W$ and $Z_P = Z_G$ are roots. This parabola is upward facing (positive coefficient of $Z_P^2$), so at $Z_P = \{Z_W,Z_P\}$ it will cross zero and go positive outside of there and be negative between those two points.
Thus the transmission of Scenario 2 will be lower only if $Z_P < Z_W$ or $Z_P > Z_G$. In other words, the transmission would be worse if you put the glass between the water and polymer or sandwich water between polymer and glass.