Oka, I will try to be didactic, succesfully or not. Don't call me heretic for recalling Schrödinger's representation, because this is just an example to ease the concepts.
People say QM is not intuitive. Well, Schrödinger's vision is intuitive enough... once you get rid of Classical prejudices. If you think of particles as waves, then things are not only intuitive, but well known results from classical optics.
The uncertainty principle already arises in classical optics, because it is inherent to Fourier's theorem. The width of the spectrum is inversely proportional to the extension of the wave pulse. This is for plain waves, but something very simmilar applies to waves whose group velocity is a half phase velocity, as in spinless QM.
Imagine an harmonic wave. You know that harmonic waves are made by only one frequency, and hence one only $k$, as $\omega=ck$. So, a typical harmonic wave is made by only one $k$. Now I ask: where is this wave? The answer is... everywhere! It is extended from minus infinite to infinite... You know its momentum $(p=\hbar k)$, but you can't tell the position: the amplitude is maximum in so many points, and since probability is proportional to $A^2$, there are so many points where the particle CAN be found with so much probability.
Now imagine a wavefunction concentrated around one point, like a gaussian centered around $x_0$. The probability of finding the particle is now clearer. The particle will be probably found around $x_0$ and it is extremely unlikely to foind it farther than $3\sigma$'s from $x_0$. So now the uncertainty in position is somehow related to $\sigma$.
So we have an example of a wavefunction that is "more localized" around a point. However, if you compute its spectrum, you will find that this function can be written as a superposition of MANY harmonic waves, each one having different frequencies and $k's$. That's Fourier theorem: a continous Square-integrable can always be written as a sum of harmonic waves. The sum might have infinite terms or even be contunous (integral), but it is after all a superposition of harmonic waves.
If that function is equivalent to a sum of harmonic waves, if you try to measure momentum, you can have as many values as different "p's" appear in that superposition. (Example: if you have a beam of red and green photons, you can measure frequency of one, and you can find a red one or a green one, but not a blue one).
So, the more localized position is, the more $k$'s it is made of. So, if you have a narrow wavefunction, you will certainly find it around $x_0$ ($\Delta x$ will be small), but momenta will be so many, so the uncertainty in momenta will be big.
The product of them will always be greater or equal than $\hbar/2$. In the case of a gaussian packet, it wiill be exactly that quantity, but that's not that normal.
Hope this helps.