In short: the spinning fields correspond to representations of the Lorentz group that we see arising for an isolated rotating body.
Internal angular momentum as Lorentz group generator
As a demonstration, consider a cloud of non-interacting particles in Minkowski space-time that you are viewing from afar. Let me use 3+1 splitted Minkowski coordinates so that I can define a Hamiltonian formalism and denote the coordinates as $t,x^i$ with $i,j,k=1,2,3$. Now each particle with mass $m_a$, $a = 1...N$ will have an on-shell 4-momentum
$$P^\alpha = \left(\sqrt{m_a - \sum (P^i_a)^2}, P^1_a,P^2_a,P^3_a\right)$$
I also define the formal 4-position as $x^\alpha_a(t) = (t,x^i_a(t))$, where $t$ is simply the value of the time coordinate used to parametrize the motion of all the particles.
This Hamiltonian system naturally has the Poisson algebra $\{x^i_a,P_{ja}\} = \delta^i_j$ for every $a$ and otherwise all the variables commute. (Note that $P_i=P^i$ in Minkowski coords.) Now define an angular momentum tensor
$$M^{\alpha \beta} = \sum_a (x^\alpha_a - x^\alpha_{\rm c.})P_a^\beta - (x^\beta_a - x^\beta_{\rm c.})P_a^\alpha $$
where all the functions on the right-hand side are given as functions of $t$, and $x^\alpha_{\rm c.} = (0,x^i_{\rm c.})$ is some referential centroid position (which is treated as Poisson-commuting with all the other variables here, for simplicity).
Finally, we define the total momentum $P^\alpha_{\rm tot.} = \sum_a P_a^\alpha$. Now we see that $P^\alpha, M^{\beta \gamma}$ fulfill the commutation relations of the generators of the Poincaré group (showing only non-zero brackets):
$$
\{M^{\alpha \beta},M^{\gamma \delta}\} = \eta^{\alpha \gamma} M^{\beta \delta} + \eta^{\beta \delta} M^{\alpha \gamma} - \eta^{\beta \gamma} M^{\alpha \delta} - \eta^{\alpha \delta} M^{\beta \gamma} \\
\{M^{\alpha \beta},P^\gamma\} = \eta^{\alpha \gamma} P^\beta - \eta^{\beta \gamma} P^\alpha
$$
where in this case you have to use the equations of motion to prove some of the identities. With a little bit of extra work you can see that these brackets will be the same in every 3+1 split so they can be considered as fully covariant.
Casimir elements and their relation to internal angular momentum
Now we see that there are a couple of Casimir invariants of the algebra such as $P_\alpha P^\alpha \equiv - m_{\rm tot}^2$ and $\mathcal{S} \equiv \sqrt{-M^{\alpha \beta}M_{\alpha \beta}/2}, \, \mathcal{S}^* = \sqrt{\epsilon_{\mu\nu\kappa\lambda}M^{\mu\nu}M^{\kappa\lambda}/2} $. The interpretation of $m_{\rm tot}$ as the total mass of the ensemble of particles (including their kinetic energy by $E=mc^2$!) is quite obvious. However, the interpretation of $\mathcal{S},\mathcal{S}^*$ is less obvious. We notice that in the 3+1 split we are using we can reparametrize the angular momentum tensor by the two 3-vectors
$$
J^i \equiv \frac{1}{2} \epsilon_{ijk} M^{jk} \\
D^i \equiv M^{0i}
$$
We naturally identify these as the mass dipole moment and angular momentum 3-vector with respect to the centroid $x^\alpha_{\rm c.}$. In terms of these 3-vectors we see that the Casimir invariants actually are
$$
\mathcal{S} = \sqrt{J^2 - D^2} \\
\mathcal{S}^* = \sqrt{2\vec{J}\cdot \vec{D}}
$$
Universality
Now, you have to trust me that such constructions are general enough so that this is the picture we get for any isolated classical system. Hence, general classical isolated bodies rotating around a center of mass will have non-zero $\mathcal{S}$ and possibly even $\mathcal{S}^*$.
Conversely, if $\mathcal{S}$ is real and positive for a massive isolated body, we see that the body will appear as having a non-zero angular momentum with respect to the centroid in every frame. Even more, since $M^{\alpha \beta}$ is the generator of Lorentz transformations and these Casimirs are all the Casimirs the algebra has, $m_{\rm tot}, \mathcal{S}, \mathcal{S}^*$ are the only properties of the ensemble that are invariant with respect to Lorentz transforms. When writing down coarse-grained (but still Lorentz-invariant) interactions of external fields with such rotating bodies, the numbers $m_{\rm tot}, \mathcal{S}, \mathcal{S}^*$ provide a universal set of parameters for the coupling.
Correspondence to irreducible representations
If you then switch to the quantum picture, you will see that the $(m,n)$ labels of the representations of the Lorentz group actually correspond roughly to the values of an alternative base of Casimir invariants
$$
m,n \sim \sqrt{\frac{\mathcal{S}^2 \pm i (\mathcal{S}^*)^2}{2 \hbar^2}}
$$ or
$$\mathcal{S} \sim \hbar \sqrt{m^2 + n^2} $$
Now it is easy to see that nonzero $m,n$, such as $m=n=1/2$ for vector fields, must, inevitably correspond to a complex analytical continuation of the notion of an isolated rotating body with nonzero "internal" angular momentum.
(In fact, the highest weight of $SO(3)$ appearing within the $(m,n)$ representation would actually be $m+n$, so one can actually state that $J$ can be understood as reaching up to $\sim \hbar (m+n)$, depending on the choice of centroid. This justifies calling the $(m,n)$ fields as "spin $m+n$".)
