Some derivations of Hesenberg's Uncertainty principle are based on the momentum of light or particles used in measuring an object's position and momentum. Does this lower limit on uncertainty describe the uncertainty in our measurements or some true uncertainty in the particle's position and momentum? If the latter is true then what does uncertainty in position and momentum really mean?
The Heisenberg Uncertainty principle does not follow from a measurement uncertainty. It arises from the probabilistic nature of quantum mechanics. In quantum mechanics a state describes some sort of probability distribution and the uncertainty principle is a condition that the standard deviation of these distributions have to satisfy. It follows from the commutation relations of canonical observables (e.g. position and momentum).
So uncertainty in momentum or position means the particle has a certain probability of being in a range that is characterized by this uncertainty (i.e. it is likely to be within this range).
In Quantum Mechanics, a state that one may observe is an eigenstate of an operator in a Hilbert Space. For example, if you wanted to predict the momentum of an electron in a definite sense, the electron's momentum would be represented as an eigenstate of the momentum operator for the Hilbert Space of the electron. This may seem very abstract, since in classical mechanics we use simple functions on Euclidean or Minkowskian space to represent things like position and momentum, but that simply doesn't work for Quantum Mechanics, since now Nature requires us to use a mathematical vector space called a Hilbert Space for us to predict her behavior as best we can.
Now, certain operators on this Hilbert space (which represent observables) may commute with eachother, in the same way two matrices may commute with eachother. But if these operators do not commute, then a theorem in linear algebra tells that it is impossible for a state in the Hilbert Space to simultaneously be an eigenstate of both those operators (these operators would be non-mutually diagonalizable, in technical lingo). Since measuring a system, say an electron's momentum, causes it to be in an eigenstate of the momentum operator, this precludes that same state from being an eigenstate to an operator that momentum does not commute with, say, the position operator. Thus, an eigenstate of the momentum operator cannot also be an eigenstate of the position operator, hence these two observables cannot be measured simultaneously to an arbitrary accuracy (again, because a measurement of a certain observable collapses a general state to an eigenstate of that observable).
On the other hand, if we have a free electron, we can know both its momentum and energy to a theoretically absolute degree, since the momentum and energy operators commute for a free electron. But the position operator for a free electron does not commute with neither of these operators, so the position cannot be known with certainty alongside these quantities: if you measure the momentum, then quickly measured its position, you would cause the electron to jump from a certain momentum eigenstate, to a different position eigenstate that is no longer an eigenstate of the momentum operator. And again, it is impossible for the eigenstate that the electron is represented by to be an eigenstate of both the position and momentum operators since they do not commute, and their lack of commutativity is a measure of the theoretical limits to our simultaneous, instantaneous knowledge of both of these quantities.