Significance of 4π in the uncertainty principle
The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" used. Without rigid definition of this quantity one ofthen just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter)
$$ \Delta x \Delta p \sim \hbar \, .$$
To make the statement more precise, one have to define what is actually meant by "uncertainty" and usually uncertainties are defined as the standard deviations.
$$
\Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\
\Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\
$$
where angle brackets $\langle \phantom{x} \rangle$ stand for the average value. And with these definitions one can indeed prove that
$$
\Delta x \Delta p \geq \hbar / 2 \, .
$$
Actually, we do even favour the usual notation for the standard deviation to avoid ambiguity
$$
\sigma_x \sigma_p \geq \hbar / 2 \, .
$$

So, from this perspective the 2 is nothing but the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the reduced Planck constant. $4 \pi$ is the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the Planck constant itself.



In fact, you can prove a more general Robertson uncertainty relation that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ 
$$
    \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|.
$$
Now, for $x$ and $p$ it is postulated by so-called canonical commutation relation that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to 
$$
    \sigma_x \sigma_p \geq \hbar / 2 \, .
$$

From that perspective, the Heisenberg uncertainty relation is equivalent to canonical commutation relation, and $\pi$ actually comes to the first one from the last one (as part of $\hbar$). So now the question really is why do we have $\pi$ in canonical commutation relation? And as user3371583 noted, it indeed has something to do with Fourier transform, but the connection of position and momentum through Fourier transform is not trivial.



Intuitively, however, you can get a sense of it by looking at the simpliest case of a one-dimensional de Broglie plane wave, in which $\pi$ factor in a connection between $x$ and $p$ naturally comes from de Broglie relation
$$
    p = \hbar k \, ,
$$
where $p$ is the momentum and $k = 2 \pi / \lambda$ is the wavenumber. So, wave function for plane wave can be written as follows
$$
    \psi(x) = \mathrm e^{\mathrm i(kx-\omega t)} \, ,
$$
and its derivative with respect to $x$ is
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i k \mathrm e^{\mathrm i(kx-\omega t)} = \mathrm i k \psi(x) \, .
$$
If we substitute then $p / \hbar$ for $k$ from de Broglie relation we get 
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i \frac{p}{\hbar} \psi(x) \, ,
$$
or
$$
    p \psi(x) = - \mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, ,
$$
which suggest that 
$$
    \hat{p} \psi(x) = -\mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, .
$$
$\hat{x}$ is simple multiplicative operator in this picture 
$$
    \hat{x} \psi(x) = x \psi(x) \, ,
$$
so, the commutator between $\hat{x}$ and $\hat{p}$ is indeed $[\hat{x},\hat{p}]=i\hbar$ and through this commutator $\pi$, in a sense, propogates to Heisenberg uncetainty relation.