Linear momentum being independent of the coordinate system "An Introduction to Mechanics" by Kleppner and Kolenkow has the following lines:

Angular momentum $\vec L$ explicitly involves the position vector $\vec r$. The value of $\vec L$ therefore depends not only on the motion of the particle, but also on its location with respect to the origin of a particular coordinate system. This is in contrast to the situation for linear momentum $\vec p$, which is independent of the coordinate system.

When the author mentions "coordinate system" in the above lines, is he referring to a coordinate system that is fixed to the earth( approximately inertial )? 
As far as I know, in Classical Mechanics, linear momentum $\vec p$ of a system moving on the earth is independent of any coordinate system that is fixed( to the earth ). If the coordinate systems are moving, different observers will record different momenta for a particular system, will they not?
 A: You are right. The linear momentum of a mass point depends, of course, on the inertial system you are using to describe the motion of the mass point. Most easily you can see this from $$\vec p =m\vec v$$ where $\vec v$ is the velocity in a reference frame $\Sigma$. Using the Galilean transform for low relative speed, in an inertial reference frame $\Sigma'$ moving with constant velocity $\vec v_{rel}$ without rotation relative to the first, the velocity of the mass point $$\vec v'=\vec v-\vec v_{rel}$$ Thus the linear momentum in the reference frame $\Sigma'$ will be $$\vec p'=\vec p -m\vec v_{rel}$$ 
A: What the book suggests is that suppose you are in an inertial reference frame i.e. either you are at rest with respect to the particle or you are moving at an uniform speed wrt to it. The former case is trivial as then the relative velocity of the particle is $0$, and thus the momentum is also $0$. Given the second case, you observe a momentum equal to the particle's mass times relative velocity ( within a Lorentz factor $\gamma$ ). 
The position of the origin of the reference frame is immaterial as it occurs nowhere in the calculations. This is what is referred to as an affine transformation, specifically translations and scalings. Whether you translate your coordinates anywhere, as long as your relative velocities stay the same, you will measure the same momentum of the particle. (But it might change if you rotate your reference frame or do something weird like being in an accelerated reference frame etc).
A: A simple view.
A coordinate system is defined in space by defining the location of (0,0,0) with respect to which one measures (x,y,z) and therefore the $\vec r$ vector. The momentum is not a function of $\vec r$ , so within an inertial system the definition is unchangeable.
Angular momentum is defined by the choice of (0,0,0) a moving mass will have a different angular momentum depending where one defines the (0,0,0) . Usually we define it on the axis of rotation because the angular momentum concept is used in rotating wheels etc. Conservation of angular momentum holds only in this case. 
But in the general definition  where the (0,0,0) can be chosen anyplace and an angular momentum defined, for example for a particle moving in as straight line versus a chosen origin, the angular momentum would be changing as the particle moves, as $\vec r$ would be changing, so no conservation law. See  also the answer here.
