The uncertainty principle should be understood as follows: The position and momentum of a particle are not well-defined at the same time. Quantum mechanically, this is expressed through the fact that the position and momentum operators don't commute: $[x,p]=i\hbar$.
The most intuitive explanation, for me, is to think about it in terms of wave-particle duality. De Broglie introduced the idea that every particle also exhibits the properties of a wave. The wavelength then determines the momentum through $$p=\frac{h}{\lambda}$$ where $\lambda$ is the De Broglie wavelength associated with the particle. However, when one thinks about a wave, it is clear that the object described by it will not be easy to ascribe a position to. In fact, one needs a specific superposition of waves to create a wave that is essentially zero everywhere except at some position $x$. However, if one creates such a wave packet, one loses information about the exact wavelength (since a wave with a single, well-defined wavelength will simply extend throughout space). So, there is an inherent limitation to knowing the wavelength (i.e. momentum) and position of a particle. On a more technical level, one could say that the uncertainty principle is simply a consequence of wave-particle duality combined with properties of the Fourier transform. The uncertainty is made precise by the famous Heisenberg uncertainty principle,
$$\sigma_x\sigma_p\geq \frac{\hbar}{2}$$
More generally, for two non-comuting observables $A$ and $B$ (represented by hermitian operators), the generalized uncertainty principle reads
$$\sigma_A^2\sigma_B^2\geq \left(\frac{1}{2i}\langle [A,B] \rangle\right)^2\ \implies \ \sigma_A\sigma_B \geq \frac{|\langle [A,B]\rangle| }{2}$$
Here, $\sigma$ denotes the standard deviation and $\langle\dots\rangle$ the expectation value. This holds at any time. Therefore, the measurement occurring right now, having occurred in the past or occurring in the future has nothing to do with it: The uncertainty principle always holds.
