Why is \$e^{-j2\pi f ~t_0} \$ instead of \$e^{-j2\pi f ~(t - t_0)} \$ for time shifting in the Fourier transform?

Only \$x(t)\$ is being shifted, not the transform. By shifting the transform as well, the shift is essentially removed.

Let \$t^{'}=t-t_0\$. Edit: [Then the OP's last equation becomes (noting that \$dt^{'}=dt\$)] $$ x(t^{'}) \xrightarrow{\mathcal{F}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft^{'}} dt^{'}=X(f) $$ not $$ X(f) e^{-j2\pi f t_0} $$

as it should be

So understand that only \$x(t)\$ is being shifted, not the transform.

Let \$t^{'}=t-t_0\$. Then dt' = dt $$ x(t^{'}) \xrightarrow{\mathcal{F}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft} dt^{'} $$

$$ X^{'}(f) = \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft} dt^{'} $$ $$ = \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi f (t^{'}+ t_{o})} dt^{'} $$ $$ = \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi f (t_{0}} e^{-j2\pi f (t^{'}} dt^{'} $$ $$ = e^{-j2\pi f t_{0}} \int_{-\infty}^{\infty} x(t^{'}) e^{-j2\pi ft^{'}} dt^{'} $$

$$ X^{'}(f)= e^{-j2\pi ft_{0}} X(f) $$ This proves the time shifting property of Fourier transform. Always remember that when you are doing fourier transform, whatever the function of t, you have to integrate from -∞ to ∞ with the multiplier \$ e^{-j2\pi ft} \$ inside the integral which is not \$ e^{-j2\pi f(t-t_{0})} \$.