How do we know that the Fourier transform of space is momentum? How do we know that the Fourier transform of real space $x$ is the momentum $p$ space or for energy and time, receptively? 
What's the mathematical process and physical logic?
 A: 
What's the mathematical process and physical logic?

The Fourier transform of position space ($\vec x$ domain) is wave number space ($\vec k$ domain).  This is an unambiguous, well understood mathematical result.
By the De Broglie hypothesis, the momentum is $\vec p = \hbar \vec k$.  This is physical hypothesis with experimental confirmation.
Although the above answers the quoted question, I suspect that you won't find it satisfying as you're looking for something 'deeper'.  In that case, think carefully about what your actual question is and post it as a separate question.
For example, "what is the physical intuition that motivates the De Broglie hypothesis"?  I haven't searched but that question may already have an answer here.
A: To quantize a classical system, start from the Poisson bracket $$\{x_i, p_j\} = \delta_{ij}.$$
This relation defines $p_i$ as the momentum canonically conjugate to $x_i$ and is equivalent to Hamilton's equations. Quantize by letting $x_i, p_j$ be Hermitian operators on a Hilbert space, with commutator $$[\hat x_i, \hat p_j] = i\delta_{ij} $$ (identity operator implicit). Let $|\vec x\rangle$ denote an eigenstate with eigenvalues $\vec x$ and $\psi(\vec x) = \langle \vec x | \psi \rangle$. Then the operator $$\hat p_i : |\psi\rangle  \mapsto -i \int d^3x' |\vec x\rangle \frac{\partial }{\partial \vec x_i} \psi(\vec x)$$
satisfies the commutation relation (and is the only such operator). With this representation of the momentum operator, the eigenvalue problem is $$\langle \vec x'| \hat p_i |\psi\rangle =\vec p_i\langle x'|\psi\rangle = \vec p_i \psi(x') = -i\int d^3\vec x\,  \delta(\vec x-\vec x') \frac{\partial}{\partial \vec x_i} \psi(\vec x)$$
or using the Dirac delta,
$$p_i \psi(x) = -i\frac{\partial}{\partial \vec x_i}\psi(x).$$
Thus, for an eigenstate of momentum, $\psi(x) \propto e^{\vec p\cdot \vec x}$. That is, the state is a plane wave.
To summarize, from the canonical Poisson brackets, quantized to the canonical commutation relations, we find that momentum eigenstates are plane waves. Thus expressing a problem in terms of momentum eigenstates is the same as expressing it in plane waves, which is precisely what the Fourier transform does.
A: In the book of Kreyszig, Introductory Functional Analysis with Applications, page 577, there is the following reasoning to associate momentum with Fourier frequency. It would be interesting to discuss if it is correct:
Using Einstein's mass-energy relation $E= mc^2$, and Planck equation $E= h\nu$ ($\nu$=frequency), we can write the momentum
$p= mc = E/c = h\nu/c = h/\lambda$
where $\lambda = c/\nu$ is the wavelength. Hence wavelength and momentum are related by
$1/\lambda = p/h$.
A plane wave of length $\lambda$ (or momentum $p$) is given by 
$\exp(2\pi i x/\lambda) = \exp(2\pi i xp/h),$
so the Fourier inversion formula
$$
\psi(x) = \int \hat\psi(\xi)\exp(2\pi i x\cdot\xi)\,d\xi
= \frac1h\,\int \hat\psi(p/h)\exp(2\pi i x\cdot p/h)\,dp
$$
gives a decomposition of $\psi(x)$ in terms of plane waves of momentum $p$.
The quantity $|\hat\psi(p/h)|$ describes the amplitude at momentum $p$.
So, if $|\psi(x)|^2dx$ is the probability density of the position in state $\psi$, then by Plancherel theorem $|\hat\psi(p/h)|^2dp/h$ should be the probability density of momentum in state $\psi$.
Finally, the mean value of momentum can be written as
$$
\mu_p=\int p|\hat\psi(p/h)|^2dp/h = h\int \xi \hat\psi(\xi){\overline{\hat\psi(\xi)}}\,d\xi= \frac{h}{2\pi i}\int \psi'\,\bar\psi=\langle P\psi,\psi\rangle,$$
if we define the momentum operator $P$ by
$$P\psi=\frac{h}{2\pi i}\frac{d}{dx}\psi.$$
A: A Fourier transform is the decomposition of a position space function into a basis of plane waves, each of which has a well defined momentum.
$$
f(x) \sim \int \text{d}p\; F(p) e^{\text{i}px}
$$
This relies on the quantum mechanical idea that waves can have a well defined momentum.
