Well, I've the impression that Griffiths leads to a lot of confusion given the many questions by readers of this book here. I, however, cannot really judge it since I've only glanced over it yet. My favorite as an introductory textbook is Sakurai. Concerning interpretation, I recommend to read Ballentine.
The (at least for me) convincing answer to the question, why the operators representing observables take the form they take in the usual representations (usually the position representation when the QM course uses the wave-mechanics approach which is pretty common for historical reasons) are symmetries. The fundamental physics nowadays is based on symmetry considerations. All starts with the principle structure of QT in terms of an observable algebra, realized as a representation on a Hilbert space. Then to fill this with concrete live for applications you have to start with the usual space-time models. To begin with, it should of course be Galilei-Newton spacetime although from a group-theoretical point of view it's a bit more complicated than Minkowski space. You meet the full case of complications, i.e., that the quantum realization of the Galilei group is both a covering group (substituting SU(2) as the representation of rotations in favor of the classical rotation group SO(3)) and a central extension (introducing mass as a central charge of the Galilei group; the mathematically possible case of 0 mass doesn't give a sensible physical model in non-relativstic QT) of the classical group.
As very first approach to the use of group theoretical arguments you can however boil down the problem to describe a particle in one-dimensional motion first, as is usually done in the hand-waving approach to wave mechanics. Roughly the argument goes as follows: You start somehow heuristically to introduce the wave function ##\psi(t,x)## as the probability amplitude and as a member of the Hilbert space of square-integrable functions of real numbers, ##\mathrm{L}^2(\mathbb{R})##. Symmetries are realized as unitary transformations on this Hilbert space (this is a bit imprecise but as a heuristic introduction it's enough to avoid confusion about subtle details which become important only later). The only symmetry of the real line as a model for space is its translation invariance (homogeneity), and you now may ask, what's the unitary transformation that describes translations, but that's very simple: Consider an observer A (Alice) who puts the origin of her coordinate at some place and another observer B (Bob) who puts the origin at the coordinate ##a## (in terms of Alice's reference frame). Then B uses ##x'=x-a## instead of Alice who uses ##x## as her coordinate for the location of the particle on the line. Now it's clear what happens: If the particle is in a (pure) state described by Alice by the wave function ##\psi(t,x)##, also for Bob it should be described by this wave function but in terms of his reference frame, ##\psi'(t,x')##. Which leads to
$$\psi'(t,x')=\psi(t,x)=\psi(t,x'-a).$$
This implies that the translation from A's to B's reference frame is described by the unitary operator
$$\hat{T}(a) \psi(t,x)=\psi(t,x-a).$$
That this is indeed a unitary transformation is trivial since for any two wave functions you have
$$\langle \hat{T}(a) \psi_1|\hat{T}(a) \psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(t,x-a) \psi_2(t,x-a) = \int_{\mathbb{R}} \mathrm{d} x \psi_1^*(t,x) \psi_2(t,x) = \langle \psi_1|\psi_2 \rangle.$$
Now in classical mechanics momentum is the generating function of the canonical transformation in Hamilton mechanics that describes spatial translations, and so we define the momentum operator in quantum theory. Since we want to have self-adjoint rather than anti-self-adjoint operators we introduce and additional factor of ##\pm \mathrm{i}## (which sign you choose is convention, and we use simply the convention used all the time from Schrödinger's seminal paper on wave mechanics on).
For an infinitesimal translation we have
$$\hat{T}(\delta a) \psi(t,x)=\psi(t,x-\delta a)=\psi(t,x)-\delta a \partial_x \psi(t,x) + \mathcal{O}(\delta a^2)=:\psi(t,x) -\mathrm{i} \delta a \hat{p} \psi(t,x)+\mathcal{O}(\delta a^2).$$
Now we have by comparison
$$\hat{p}=-\mathrm{i} \partial_x.$$
Now all this is in natural units, where ##\hbar=1##. If you want conventional units you have to introduce the appropriate power of ##\hbar## which by dimensional reasoning leads to
$$\hat{p}=-\mathrm{i} \hbar \partial_x.$$
This establishes why ##\hat{p}## should take the form it does from symmetry principles taken right from the classical theory (in terms of Hamilton mechanics).
Now we also can state another important piece of mathematics. The whole formalism can be just based on purely algebraic arguments using the symmetries. The infinitesimal translations can be defined by the Lie algebra, i.e., the algebra of linear operators in Hilbert space with the composition as product. For the Lie algebra you only need the commutators, and the only non-trivial commutator of the algebra of a quantized particle along a line is that between position and momentum. Using the above result (again with ##\hbar=1## for convenience) we find
$$\hat{x} \hat{p} \psi(t,x)=x (-\mathrm{i} \partial_x) \psi(t,x)$$
and
$$\hat{p} \hat{x} \psi(t,x)=(-\mathrm{i} \partial x) x \psi(t,x)=x(-\mathrm{i} \partial_x) \psi(t,x)-\mathrm{i} \psi(t,x),$$
which by subtraction from the previous equation leads to
$$(\hat{x} \hat{p}-\hat{p} \hat{x}) \psi(t,x)=[\hat{x},\hat{p}] \psi(t,x)=\mathrm{i} \psi(t,x) \; \Rightarrow \; [\hat{x},\hat{p}] =\mathrm{i} \hat{1}.$$
You can build all the rules concerning operators of a particle moving along a line from this commutation relation!
