Fourier transform of the components of bandpass signals
If you multiply \$s(t)\$ with \$2\cos(2\pi f_c t)\$ and with \$-2\sin(2\pi f_c t)\$, respectively (i.e. you are actually demodulating the signal), you get:

$$s(t)\cdot2\cos(2\pi f_c t)=2s_I(t)\cos^2(2\pi f_c t)-2s_Q(t)\sin(2\pi f_c t)\cos(2\pi f_c t)=\\
=s_I(t)+[s_I(t)\cos(2\cdot2\pi f_c t)-s_Q(t)\sin(2\cdot2\pi f_c t)]$$

and

$$-s(t)\cdot2\sin(2\pi f_c t)=-2s_I(t)\cos(2\pi f_c t)\sin(2\pi f_c t)+2s_Q(t)\sin^2(2\pi f_c t)=\\
=s_Q(t)-[s_I(t)\sin(2\cdot2\pi f_c t)+s_Q(t)\cos(2\cdot2\pi f_c t)]$$

Note that the terms in brackets are centered at twice the carrier frequency \$f_c\$. So in the band \$-W&lt;f&lt;W\$ (i.e. by low-pass filtering) we get

$$s(t)\cdot2\cos(2\pi f_c t)=s_I(t)\\
-s(t)\cdot2\sin(2\pi f_c t)=s_Q(t)$$

If you take the Fourier transform of these two equations you obtain the Fourier relations stated in your question.