So what are spin-networks? Briefly, they are graphs with representations ("spins") of some gauge group (generally SU(2) or SL(2,C) in LQG) living on each edge. At each non-trivial vertex, one has three or more edges meeting up. What is the simplest purpose of the intertwiner? It is to ensure that angular momentum is conserved at each vertex. For the case of four-valent edge we have four spins: $(j_1,j_2,j_3,j_4)$. There is a simple visual picture of the intertwiner in this case.
Picture a tetrahedron enclosing the given vertex, such that each edge pierces precisely one face of the tetrahedron. Now, the natural prescription for what happens when a surface is punctured by a spin is to associate the Casimir of that spin $ \mathbf{J}^2 $ with the puncture. The Casimir for spin $j$ has eigenvalues $ j (j+1) $. You can also see these as energy eigenvalues for the quantum rotor model. These eigenvalues are identified with the area associated with a puncture.
In order for the said edges and vertices to correspond to a consistent geometry it is important that certain constraints be satisfied. For instance, for a triangle we require that the edge lengths satisfy the triangle inequality $ a + b \lt c $ and the angles should add up to $ \angle a + \angle b + \angle c = \kappa \pi$, with $\kappa = 1$ if the triangle is embedded in a flat space and $\kappa \ne 1$ denoting the deviation of the space from zero curvature (positively or negatively curved).
In a similar manner, for a classical tetrahedron, now it is the sums of the areas of the faces which should satisfy "closure" constraints. For a quantum tetrahedron these constraints translate into relations between the operators $j_i$ which endow the faces with area.
Now for a triangle giving its three edge lengths $(a,b,c)$ completely fixes the angles and there is no more freedom. However, specifying all four areas of a tetrahedron does not fix all the freedom. The tetrahedron can still be bent and distorted in ways that preserve the closure constraints (not so for a triangle!). These are the physical degrees of freedom that an intertwiner possesses - the various shapes that are consistent with a tetrahedron with face areas given by the spins, or more generally a polyhedron for n-valent edges.
Some of the key players in this arena include, among others, Laurent Friedel, Eugenio Bianchi, E. Magliaro, C. Perini, F. Conrady, J. Engle, Rovelli, R. Pereira, K. Krasnov and Etera Livine.
I hope this provides some intuition for these structures. Also, I should add, that at present I am working on a review article on LQG for and by "the bewildered". I reserve the right to use any or all of the contents of my answers to this and other questions on physics.se in said work, with proper acknowledgements to all who contribute with questions and comments. This legalese is necessary so nobody comes after me with a bullsh*t plagiarism charge when my article does appear :P
Best Answer
Reverse the burden: Why should there be a unique quantization method? The classical theory is a limit of the quantum theory, why should this limit be reversible? It's like asking thermodynamics to be recoverable from a zero-temperature (or any other) limit, or the $\mathbb{R}^{6N}$ phase space dynamics to be recoverable from the thermodynamic limit $N\to\infty$. There's no reason to expect the full theory to be encoded in one of its limits, in fact no reason for us to expect the existence of a quantization method at all, let alone a unique one.
Quantization is obstructed: A "quantization" is supposed to be an assignment of Hermitian operators on a Hilbert space to classical observables on phase space, i.e. a map $f(x,p)\mapsto \hat{f}$. The Groenewold-van Hove theorem says that there is no such map such that
meaning every quantization method must drop some of these assumptions, and it usually does not suffice to only drop the fourth. Canonical quantization usually assumes all of this works anyway, and when it goes wrong it's fixed ad hoc. Deformation quantization drops the fourth property and make sthe second hold only up to terms of order $\hbar^2$, geometric quantization instead restricts the allowed inputs $f$ to the quantization map and drops the fourth property.
Therefore, you naturally get different quantization methods depending on which assumptions you're willing to sacrifice. As a matter of fact, it is not known for any of the quantization methods whether they are "equivalent" in a fully general setting. Additionally, this does not even begin to cover all possible "quantizations", since e.g. the path integral formalism is not a map $f\mapsto \hat{f}$. Alas, it is not strictly known whether it is truly equivalent to the operator formalism, but most known cases seem to don't differ between the two formalisms. For a longer discussion of that point, see this question.