Given $H(x)$ is the Heaviside Theta, the tables give the following Fourier transforms for it:


$$ H(x+a)\to -PV\frac{i e^{i a w} }{w}+\pi \delta (w)$$


while from Sokhotski–Plemelj theorem it follows (seemingly) that


$$ H(x+a)\to~PV\frac{-i}{w} +\pi\delta(w)+ |a|e^{iax/2}\operatorname {sinc} \left({\frac {w a }{2\pi}}\right)$$


The first result seems questionable to me because it does not depend on $a$ at $w=0$. But this should not be the case because $H(x+a)$ can be represented as a sum of $H(x)$ and the $rect$ function, and the Fourier transform of the later depends on $a$.



Which Fourier transform is the correct one?
application of the Sokhotski–Plemelj formula does in fact give the first of the two expressions:


$$\int\_{-\infty}^\infty H(x+a)e^{-ix w}dx=\int\_{-a}^\infty e^{-i xw}dx=e^{iwa}\int\_0^\infty e^{-ix w}dx$$
$$=e^{iwa}\left(\text{PV}\frac{-i}{w}+\pi\delta(w)\right)=-\text{PV}\frac{ie^{i aw}}{w}+\pi\delta(w)\qquad(\ast)$$


where PV indicates the principal value; so this is the first of the two formulas in the OP; it may equivalently be written as
$$(\ast)=-\text{PV}\,\frac{i\cos aw}{w}+\pi\delta(w)+a\,\text{sinc}\,aw$$
which is different from the second formula




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the OP asks whether I can reach the same result by splitting $H(x+a)$ into $H(x)$ plus a rect function:


Fourier transform of $H(x)$: $\int\_{0}^\infty e^{-ixw}dx=-\text{PV}\,\frac{i}{w}+\pi\delta(w)$


Fourier transform of rect: $\int\_{-a}^0 e^{-ixw}dx=\frac{i}{w}\left(1-e^{iwa}\right)=\text{PV}\,\frac{i}{w}(1-\cos aw)+a\,\text{sinc}\,aw$  

since $w^{-1}(1-\cos aw)=\text{PV}\,w^{-1}(1-\cos aw)$


summing up gives the same result as above, $(\ast)=-\text{PV}\,\frac{i\cos aw}{w}+\pi\delta(w)+a\,\text{sinc}\,aw$


