Why is it that the Laplace transform of $f(t)$ evaluated at $s = j\omega$ is not the Fourier transform

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Many sources state that the Laplace transform is simply the Fourier transform where $\text{Re}(s) =0 \implies s = j\omega$

However, why is it that substituting $s = j\omega$ into Laplace transforms of a function $f(t)$ do not yield the corresponding Fourier transform (for large number of transforms)?

For example,

The laplace transform of $\sin(\omega_o t)$ is $$\dfrac{\omega_o}{s^2 + \omega_o^2}$$

(According to source)

Where as the fourier transform yields:

$$i\pi (\delta(\omega+\omega_o) - \delta(\omega-\omega_o))$$

(According to source)

But evaluating $$\dfrac{\omega_o}{s^2 + \omega_o^2}|_{s = i\omega} = \dfrac{\omega_o}{-\omega^2 + \omega_o^2}$$

Quite the stark contrast.

When does the Fourier transform and the Laplace transform coincide for $s = j\omega$

Why is that sometimes they do not coincide?