With the photos, I'd venture that you have stray capacitances all over your circuit because it is on a solderless breadboard. You shouldn't be doing high frequency work like this on one of those. Of course, this doesn't rule out parasitic effects addressed in the comments to your question, but there is a gorilla in the room, and it's the breadboard.
Your question has a simple answer, but allow me explain a few things that will help you understand how to design Direct Coupled filters.
1) Cohn described the four inverter circuits I show on the top row. These circuits contain negative component values which, as you pointed out, are absorbed into the tank elements.
It is clear however that a negative capacitor generates inductive reactance and a negative inductor generates capacitive reactance. So all of Cohn's inverters have an alternate equivalent that I show in the 2nd row, where the negative components are replaced with positive valued components, which can also be absorbed into the tank circuit.
2) These inverter circuits can be thought of as impedance matching circuits. As a simple example, if you are designing an inverter to go between 2 tanks with characteristic impedances of 100 and 200 Ohms, then a quarter wave transmission line with a characteristic impedance of sqrt(100*200) can be used as an inverter.
With this in mind, it is possible to generate many possible inverter topologies.
3) If the tank circuits on the ends have characteristic impedances equal to the source and load resistances respectively, then the two inverters on the ends are not needed. But this isn't usually the case. You can force this to happen, but it doesn't usually generate a desirable circuit in terms of component values and component Q requirements.
So, one possibility for the end inverters is to use one of the inverters shown in the 2nd row.
It is more common however, to use a 2 element matching circuit to get from the end tank impedance to the source (or load) impedance. These are designed with standard matching techniques, using either formulas or a Smith Chart.
4) You should remember that impedance inverters are equivalent circuits at the design frequency only. At all other frequencies they only approximate the desired response. Consequently, this design method can only be used for narrow band filters, and then, the overall response might not be what you were expecting, as shown below.
Since we usually want a narrow band response however, this design method is widely used by RF engineers and has proven to be quite powerful because of the large number of ways (some examples) the various tank, inverter, and matching topologies can be combined to form a desirable response.
Best Answer
Almost certainly your inductor core is saturating. The currents into the 6.5 uF will be quite high and this is the reason. You have basically a series tuned circuit formed by 2x 470uH and a 6.5uF capacitor.
This will have a resonant frequency of: -
f = \$\dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{0.001\times 6.5\times 10^{-6}}}\$ =1974 Hz.
Ideally, you want the resonant frequency to be logarithmically half way between 50 Hz and 16 kHz and this would be more like 894 Hz so maybe you are a little close to the switching frequency. You could probably go lower - maybe towards 300Hz to improve this but, all the time the inductor or the cap is getting bigger.
Basically if you could raise your PWM frequency by 3 or 4 times you'd see an improvement I reckon.