Thin film interference is typically used to create anti-reflection coatings. In this situation (unless the angles allow total internal reflection), there are not multiple reflections. The number of reflections is equal to the number of interfaces.
Also, the ray actually interferes with itself. At each interface, a portion of the ray is transmitted and a portion is reflected (I'm ignoring losses like absorption and diffuse reflectance). The ratio in intensity between reflected and transmitted portions shows the percentages. The intensity of the ray is equal to the sum of its transmitted and reflected portions.
Suppose that light traveling through air encounters a material with index of refraction $n_m$ The transmission of light through the material can be described as
$$I_0 = I_{R} + I_{T}$$
$$I_{T} = I_0 - I_{R}$$
where $I_{R}$ and $I_{T}$ are the reflected and transmitted intensities.
Suppose now, a different material of given thickness and different index of refraction is deposited on top of the original material (anti-reflection coating AR), then the transmission of light through this material can be described as:
$$ I_0 = I_{AR(reflected)} + I_{AR(transmitted)}$$ and then that light is reflected from and transmitted through the material/AR boundary $$I_{AR(transmitted)} = I^{'}_{AR(reflected)} + I^{'}_T
$$
Finally
$$I_0 = I_{AR(reflected)} + I^{'}_{AR(reflected)} + I^{'}_T$$
Clearly if the two reflected components could cancel each other out$$I_{AR(reflected)} + I^{'}_{AR(reflected)} = 0$$ then 100% transmission would be possible $$I^{'}_T = I_0$$
This is what AR coating design takes advantage of and how the destructive interference that cancels these reflected waves increases transmission.