Rarita-Schwinger spin projection operators Chapter 2 of the paper Symmetry of massive Rarita-Schwinger fields by T. Pilling mentions "the usual" spin projection operators. However, to me, they are not usual and I struggle with intuition and notation.
I understand that we find the correct Lorentz representation of the RS vector-spinor by taking the tensor product of a bispinor and vector representation (eq 1 in the paper):
$$\left[(1/2,0)\oplus(0,1/2)\right] \otimes (1/2, 1/2) \quad\quad\quad$$
$$\quad\quad\quad = (1, 1/2) \oplus (1/2, 1) \oplus (1/2, 0) \oplus (0, 1/2) \tag{1}$$
My main question is about the later mentioned projection operators. Clearly a RS-spinor $\psi_\mu$ has a mixture of 3/2 and 1/2 degrees of freedom. We are not interested in the 1/2 background, so we need a projection operator $P^{3/2}$ to get rid of them. Fine. Likewise, I can define an operator $P^{1/2}$ to get the spin-1/2 background. That's just a mathematical excercise. Now, what are the extra indices at $P^{1/2}_{11}$, $P^{1/2}_{12}$, $P^{1/2}_{21}$ and $P^{1/2}_{22}$? In the paper they are described as

the individual projection operators for the two different spin-1/2 components of the
Rarita-Schwinger field

This is where I'm lost. What am I projecting? For what reason? Can someone explain this to me and give me some intution?
edit:
To clarify, consider the Projector
$$P_x = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$
The intution here is, that is projects the x-component of a three vector. Using this analogy, what is $P^{1/2}_{11}$ projecting?
 A: It is apparent that the clear and explicit formal paper you are reading is not helping you systematize what you may have learned in your QFT theoretical course (this is distinctly not experimental HEP!), namely, the basic facts summed up in the paragraph following equation (12): at the end of the day, the independent conditions $\gamma \cdot \psi$ and $p\cdot \psi$ have no Lorentz indices anymore, and so they Lorentz- transform like plain spinors, which you know represent spin 1/2 s:

...as can be seen by multiplying the first equation in (10) on the left by $γ_μ$ and using the second equation. The condition $γ_\mu^{AB}ψ^μ_B = 0$ (where we now explicitly write the spinor indices A,B) represents a constraint equation for each value of the spin index A, whereas the condition $∂_μ ψ_B^μ = 0$ is an equation of motion for the spinor components $ψ_B^0 $. However, the Dirac equation (10) also gives an equation of
motion for the same spinor components and when taken together, these result in
another set of constraints. In four space–time dimensions, these two sets of equations each constitute four constraints; and serve to remove eight components of the 16 spinor components  of the  vector-spinor $ψ_A^μ$, leaving 2(2s + 1) = 8 physical degrees of freedom as  required for a massive spin s = 3/2 particle.

That is, of the original 16 d.o.f. of the reducible field $\psi^\mu$, you prune out 4 d.o.f. by the first condition, and another 4 by the second, being left with 8 for the spin-3/2 block: a parity doublet of the 4 spin states of the spin quartet. (Remember: we have not gauged out the intermediate helicity $\pm 1/2$ states, since susy-gauge-invariance has not been assumed, this not being supergravity; it might as well be a Δ baryon.)
Now the author spends quite some time giving you a formal (Ogievetskian) implementation of these maneuvers, through the projector $(P^{3/2})^2=P^{3/2} $, the only operator you really need to appreciate,
$$
P^{3/2}_{\mu\nu}= \frac{1}{6p^2} \bigl (
4(p^2 g_{\mu\nu} - p^\mu p^\nu) \\ -p^2[\gamma_\mu,\gamma_\nu ]+p^\mu p^\kappa [\gamma_\kappa,\gamma_\nu] - p^\nu p^\kappa [\gamma_\kappa,\gamma_\mu] \bigr ) \tag{7.1} ~~~~\leadsto  \\
\gamma^\mu P^{3/2}_{\mu\nu}=0=P^{3/2}_{\mu\nu} \gamma^\nu = 0=p^\mu P^{3/2}_{\mu\nu}=0=P^{3/2}_{\mu\nu} p^\nu.
$$
As a result, after the two piecemeal projections of the two spin 1/2 s, the pure spin 3/2 field you only need consider is $\tilde \psi^\mu = P^{3/2}_{\mu\nu}  \psi^\nu$, which, now, automatically satisfies the conditions, by above!

This is where I'm lost. What am I projecting? For what reason? Can someone explain this to me and give me some intuition?

You are projecting the spin 3/2 piece $P^{3/2}_{\mu\nu}  \psi^\nu$ of the reducible field $\psi^\mu$ so you don't get distracted by the two irrelevant spin 1/2 spinors unfortunately also packaged in the vector-spinor; a demonstrably real risk, ipso facto. This is the intuition.
If, despite this, you really wished to  look in the dross pile for the spin 1/2 pieces you discarded in two stages, 1.13 of van Nieuwenhuizen's review will furnish an overcomplete analysis of the polarization components of the two spin 1/2 s. In (1), $P^{1/2}_{22}$ of course projects onto the spin 1/2 of $(0,1/2)\oplus(1/2,0)$, while $P^{1/2}_{11}$ onto the remnant spin 1/2 spinor in the incompletely pruned $(1,1/2)\oplus (1/2,1)$, which still has 12, not 8, d.o.f.
For the simpler, massless, case, look at section II of van Nieuwenhuizen, P., Sterman, G., & Townsend, P. K. (1978): Unitary Ward identities and BRS invariance Phys Rev, 17 1501.
A: After hours of more research, I came to the following conclusion:
A representation of the inhomogeneous Lorentz group (ILG) has a spin space $S=\{s_\mathrm{max}, s_\mathrm{max}-1, ... , s_\mathrm{min}\}$, where each spin has a multiplicity $n(s)$, where $n(s_\mathrm{max})=1$.
A Lorentz vector has a spin decomposition of $1\oplus0$ and hence for a vector-spinor we find
$$\mathrm{spin}\;\left( A_\mu \otimes \psi\right) = (1\oplus 0) \;\otimes\; 1/2 = 3/2 \oplus 1/2 \oplus 1/2\;.$$
This means $n(3/2)=1$ and $n(1/2)=2$. As mentioned in the question, for a spin-3/2 particle, we do not care about the lower spin background. This demands projection operators, such as $P(3/2)$ and $P(1/2)$.
Now the question was, what are the indices at $P_{ij}(1/2)$?
To answer this, we will use a secondary notation. Let $P(s)$ be a projection operator, that selects the spin-$s$ representation of a wave function. Since $n(s_\mathrm{max})=1$, $P(s_\mathrm{max})$ always selects the irreducible representation, but if $n(s)>1$ we only find a reducible representation. Therefore, we introduce an additional projection operator $D_i(s)$ with $i = 1, ... , n(s)$ which precisely finds the irreducible representations. This means $P(s) = \bigoplus \limits_{i=1}^{n(s)} D_i(s)$.
Combining this with the above notation, $P_{ii}(s)=D_i(s)$. So the projection operators with two equal indices, find the irreducible representations of spin-$s$.
Finally, to switch between irreducible representations, we can introduce the transition (not projection!) operators $P_{ij}$ with $i\neq j$. They have the following properties:
$$P_{21}(s) P_{11}(s)P_{12}(s) = P_{22}(s)$$
$$P_{12}(s) P_{22}(s) P_{21}(s) = P_{11}(s)$$
For a more general description, see the above paper.
Further reading:
A. Aurilia and H. Umezawa, Theory of High Spin Fields, Phys. Rev. 182
Frits A. Berends et al., On Field Theory for Massive and Massless Spin 5/2 Particles, Nucl.Phys.B
