$F(x) = 0$ for $-2 \pi / k < t < 0;$
$F(x) = sin(kx)$ for $0 < t < 2 \pi / k$
I am really confused in how to find the Fourier series of this function, supposing it is periodic with period $4 \pi /k$
$$\int _{-2 \pi /k}^{2 \pi /k} F(x')cos(nkx')dx' = \int _{0}^{2 \pi /k} sin(kx')cos(nkx')dx'=0$$ $$\int _{-2 \pi /k}^{2 \pi /k} F(x')sin(nkx')dx' = \int _{0}^{2 \pi /k} sin(kx')sin(nkx')dx'=0$$
Since both integral are zero, the coefficients in the expansion would be zero too, but this is absurd. Where is my error?
You may have misremembered the formula.
In fact, the Fourier series of a function $f$ with period $2l$ is
$$a_n=\frac{1}{l}\int_{-l}^l f(x)\cos\frac{n\pi}{l}xdx$$
$$b_n=\frac{1}{l}\int_{-l}^l f(x)\sin\frac{n\pi}{l}xdx$$
$$f\sim\frac{a_0}{2}+\sum_{n=1}^\infty(a_n\cos\frac{n\pi}{l}x+b_n\sin\frac{n\pi}{l}x)$$
So in your problem,
$$a_n=\frac{k}{2\pi}\int_0^{2\pi/k}\sin(kx)\cos(nkx/2)dx=\frac{1}{2\pi}\int_0^{2\pi}\sin(t)\cos(nt/2)dt=\begin{cases} \frac{-4}{\pi(n^2-4)} &,n \text{ odd}\\ 0 &,n \text{ even}\\ \end{cases}$$
$$b_n=\frac{k}{2\pi}\int_0^{2\pi/k}\sin(kx)\sin(nkx/2)dx=\frac{1}{2\pi}\int_0^{2\pi}\sin(t)\sin(nt/2)dt=\begin{cases} \frac{1}{2} &,n=2\\ 0 &,\text{otherwise}\\ \end{cases}$$
$$f\sim \frac{1}{2}\sin(kx)+\sum_{n=1}^\infty \frac{-4}{\pi(4n^2-4n-3)}\cos((n-\frac{1}{2})kx)$$