Problem with the Fourier transform of |sin(x)|

You can obtain a simple expression using the convolution theorem. A single cycle of the waveform is given by

sig = Sin[t] UnitBox[t/π - 1/2]

This has a relatively simple Fourier Transform

spec = FullSimplify[FourierTransform[sig, t, ω]]
(* -((1 + E^(I π ω))/(Sqrt[2 π] (-1 + ω^2))) *)

We can convert our single cycle into a periodic signal by convolution with a Dirac Comb. This transforms to another Dirac comb

combspec = FourierTransform[DiracComb[t/π], t, ω]
(* Sqrt[π/2] DiracComb[ω/2] *)

By the convolution theorem (this probably needs multiply by a something like 2π), the Fourier transform of periodic signal is given by

Assuming[ω/2 ∈ Integers, FullSimplify[combspec*spec]]
(* DiracComb[ω/2]/(1 - ω^2) *)

Note that we can simplify the factor in front of the comb, as its value only needs to be correct for cases where the comb is non-zero.

Abs isn't analytic, and using it here seems to lead FourierTransform to get lost in the complex plane. Use RealAbs:

FourierTransform[RealAbs[Sin[t]], t, w]

yielding:

Since the function is periodic, its transform is a train of delta functions. Something like FourierSeries might be more illuminating here. Maybe something like:

ComplexExpand[FourierSeries[RealAbs[Sin[t]], t, 10]]

yielding:

is what you want.