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.
The displacement of each particle on one side of the boundary with be equal to the sum of the displacement due to the incident wave $\vec y_{\rm i}$ and the reflected wave $\vec y_{\rm r}$ and particles on the other side of the boundary will have a displacement due the transmitted wave $\vec y_{\rm t}$.
If at the boundary $\vec y_{\rm i}+\vec y_{\rm r}\ne \vec y_{\rm t}$ there would be a gap between the particle immediately one side of the boundary and the particle on the other side of the boundary which is not possible.
The slope condition is a much more subtle condition and it might help to see how the wave equation is derived for transverse waves on a string.
For a more detailed explanation please have a look at one of the many derivations that you can find on the Internet.
The diagram below shows a section of a string $AB$ of mass per unit length $\mu$ with one end at position $x$ displaced from the equilibrium position by $y_1$ and the other end at position $x +\Delta x$ displaced from the equilibrium position by $y_2$.
$T$ as the tension in the string.
The diagram is greatly exaggerated in that the angles are small and their difference is small.
The net force in the y-direction on the element $AB$ is $T \sin \theta_2 - T \sin \theta_1$.
$T \sin \theta_2$ is the gradient of the string at $x+\delta x$ which I shall write as $\left(\dfrac{dy}{dx}\right)_{x+\Delta x}$ and similarly the gradient of the string at the other end is $\left(\dfrac{dy}{dx}\right)_x$
So the net force on the element is
$$T\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - T\left(\dfrac{dy}{dx}\right)_x = T\Delta x \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$
Now using Newton's second law the acceleration $a$ in the y-direction can be found.
$$\mu \Delta x \,a = T\Delta x \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$
$$\Rightarrow a = \dfrac{T}{\mu} \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$
So the acceleration of the string depends on the rate of change of the gradient of the string.
For successive elements of the string the gradient and hence the acceleration of the string will change.
However imagine that the element was at a boundary and it had a kink in it which is saying that the gradient changes in a discontinuous manner as one crosses the kink.
This would mean that the gradient of the string on one side of the kink would be very different from the gradient of the string on the other side of the kink.
Now the element of the string with the kink in it would be subjected to a very much larger acceleration than elements on either side of it.
This is not possible and so there can be no kink in the string and at a boundary the gradient of the string must be the same on both sides.
Best Answer
The two expressions look the same to me, with the difference being on the label placed on each medium.
You should focus on the meaning of transmission an reflection coefficient. That is, for a plane wave with unitary amplitude how much of it transfers to the other medium and how much gets reflected. Once you get that, you write down the equations and apply the relations between particle velocity and pressure (given by the characteristic impedance) and boundary conditions.
This is pretty much what is done in this section of Kinsler's book.
Edit: 2019-06-12
Yes, it is possible. The amplitude can go up or down, depending on the impedance contrast. Although, I think I understand what the problem is. You might be thinking that this implies that the power is increasing, but that's not the case. For that, you should look at the intensity (or power) transmission coefficients given by (6.2.10 and 6.2.11):
$$R_I =\left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2\, ,$$
and
$$T_I = \frac{4 Z_2 Z_1}{(Z_2 + Z_1)^2}\, .$$
Now, $T_I$ cannot increase from one medium to the other but $T$ can. Let us consider $Z_1 = 1$ and $Z_2 = 4$, in that case, we have
$$R = \frac{3}{5}\, ,\quad T = \frac{8}{5}\, ,$$
but,
$$R_I = \frac{9}{25}\, ,\quad T_I = \frac{16}{25}\, .$$
If we change the roles, $Z_1=4$ and $Z_2=1$, we get
$$R = \frac{-3}{5}\, ,\quad T = \frac{2}{5}\, ,$$
but,
$$R_I = \frac{9}{25}\, ,\quad T_I = \frac{16}{25}\, .$$
In the second case, we should interpret the negative sign in the reflection coefficient as a phase change of $\pi$ radians.
Notice that in both cases, the transmitted "energy" is less than 1. But this information is not encoded in the transmission coefficient itself, but in $T_I$. This makes total sense since the impedance is telling us how "easy" is to move the fluid for a given pressure.