Spectrum analysis of a function
The phase term of your outermost cosine function is time-varying: $\phi(t):=t a (1+b \cos(ct))$. The instantaneous frequency is given by $\frac{\partial \phi}{\partial t} = a+ab \cos(ct) - abct \sin(ct).$ In particular, in a tiny neighborhood around $t=0$ we can argue that $f(t)$ has only one frequency component $a+ab$. At large values of $t$, the $abct\sin(ct)$ term takes control.

Notice that the frequency varies with time. This makes it difficult to intuitively interpret the Fourier transform which assumes that all frequencies exist for all times. For such signals, it may be better to use time-frequency analysis by looking at the spectra of windowed segments of $f(t)$ around different values of $t$.