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The answer is no, at least for the first two interpretations of your question that I see: I assume that $\Omega$ is a $m$-dimensional compact submanifold of $\mathbb R^n$ with induced Riemanian Riemannian metric $g$ and volume form $\operatorname{vol}(g)$.

1. Interpretation: $f$ is a distribution of compact support in $\mathbb R^n$ acting on test function $\phi$ by $\langle f,\phi\rangle = \int_{\Omega} \phi.f \operatorname{vol}(g)$. Then Fourier transform of $f$ is a tempered distribution, not rapidly decreasing since $f$ has singular support.

2. Interpretation: $f$ is a cocurrent. It acts on smooth $m$-forms $\omega$ with compact support as $\langle f,\omega\rangle = \int_{\Omega} f.\omega$. Here your Sobolev index has to be $k> \dim(\Omega)/2 +1$ so that $f.\omega$ is at least a $C^1$ form.

3. Interpretation: If $\Omega$ is a symmetric space of certain type isometrically embedded in $\mathbb R^n$ there might be a (spherical) Fourier transform for functions on $\Omega$ itself. This is quite subtle.

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The answer is no, at least for the first two interpretations of your question that I see: I assume that $\Omega$ is a $m$-dimensional compact submanifold of $\mathbb R^n$ with induced Riemanian metric $g$ and volume form $\operatorname{vol}(g)$.

1. Interpretation: $f$ is a distribution of compact support in $\mathbb R^n$ acting on test function $\phi$ by $\langle f,\phi\rangle = \int_{\Omega} \phi.f \operatorname{vol}(g)$. Then Fourier transform of $f$ is a tempered distribution, not rapidly decreasing since $f$ has singular support.

2. Interpretation: $f$ is a cocurrent. It acts on smooth $m$-forms $\omega$ with compact support as $\langle f,\omega\rangle = \int_{\Omega} f.\omega$. Here your Sobolev index has to be $k> \dim(\Omega)/2 +1$ so that $f.\omega$ is at least a $C^1$ form.

3. Interpretation: If $\Omega$ is a symmetric space of certain type isometrically embedded in $\mathbb R^n$ there might be a (spherical) Fourier transform for functions on $\Omega$ itself. This is quite subtle.