Does it matter if Poincaré transformations are just coordinate changes? In a previous question of mine it was established that special relativity may be formulated as the theory of a (smooth) Lorentz manifold $(M,g)$ that is diffeomorphic to (the standard diff. structure of) $\mathbb{R}^4$ where $g$ is a flat metric of signature $(+---)$. In this case we have a representation of the group $SO(1,3)$ on each tangent space $T_pM$, always preserving $g$. Furthermore the group  $$\text{Isom}(M)=\{\phi:M\to M|\ \phi \text{ is a diffeomorphism and } \phi_*g=g\}$$ of isometries of $M$ is isomorphic to the Poincaré group. 
The continuation of this question is as follows:
If one defines a general field to be some section of a (possibly complex) vector bundle over $M$, why does one expect that vector bundle to carry a representation of the Poincaré group, or some subgroup? This as opposed to the obvious claim that the coordinate representation of such a bundle section can be transformed into the representation with respect to any other coordinate system. In particular I want to know if there is a meaningful difference between considering the Poincaré transformations as coordinate changes and considering them as more meaningful transformations, under the actions of which we would expect things like actions and fundamental equations of motion to be invariant. To illustrate what I mean: If $x,y:M\to\mathbb{R}^4$ are global charts of $M$ such that $x\circ y^{-1}$ is actually a Poincaré transformation then transforming the coordinate representations of the same field with respect to $x$ to the representation with respect to $y$ will involve Lorentz matrices. 
 A: Consider a vector bundle $E \overset{\pi}{\rightarrow} M$. In your question, you construct a diffeomorphism $\phi=x\circ y^{-1}:M\rightarrow M$ out of two global charts $x, y:M\rightarrow \mathbb{R}^4$. To me, it seems that you say that $\phi$ defines a transformation of the space of sections:
\begin{align}
\Phi:\rm{Diff}(M)\times\Gamma(M,E)&\rightarrow \Gamma(M,E) \\
(\phi, X) &\mapsto (p \mapsto X_{\phi(p)})
\end{align}
asuming that we are identifying of all the fibers $\pi^{-1}(p)$ for $p\in M$ using the flat connection associated to the flat metric $g$. Of course this defines some action of the Poincaré group.
However, for some suitable bundle a non-equivalent action of the Poincaré group can be defined. Consider a representation of the Lorentz group $R:SO(1,3)\rightarrow \rm{End}(V)$ and suppose that the fibers of our bundle are isomorphic to (and identified with) $V$. The action over $\Gamma(M,E)$ of the Poincaré group that we want is
\begin{align}
\Psi_R: \rm{Iso}(M)\times\Gamma(M,E )&\rightarrow \Gamma(M,E) \\
(\phi, X) &\mapsto (p \mapsto R(d\phi) X_{\phi(p)})
\end{align}
This is the action that is used in physics. It includes as a particular case the restriction of $\Phi$ from $\rm{Diff}(M)$ to $\rm{Iso}(M)$. This can be seen substituting $R$ by the tensor product of the trivial representation with itself. $\Psi_R$ is the action that involves the Lorentz matrices, but only when $R$ is the fundamental representation.
Now, it is true that for every Poincaré transformation $\phi$ there's an associated coordinate transformation $C^\infty(M,\mathbb{R}^4)\rightarrow
C^\infty(M,\mathbb{R}^4):x\mapsto \phi\circ x$. There's also a "trivial" action of every Poincaré transformation over the space of sections: $\Phi$, but it's not the one used in physics and it doesn't involve Lorentz matrices. For the action used in physics we need some extra information, namely the representation $R$.
As for the reason why such a non-trivial action of the Poincaré group is used in physics, in special relativity the space-time symmetry group is $\rm{Iso}(M)\simeq ISO(1,3)$ (the reason for this is, if you want, experimental evidence) and we want to know how the relevant objects in the theory transform under it. In relativistic field theory, those objects are the fields: sections of the vector bundles equipped with some $\Psi_R$.

In general, the statement that the action $S:\Gamma(M,E)\rightarrow \mathbb{R}$ is invariant under some transformation $T:\Gamma(M,E)\rightarrow\Gamma(M,E)$ means that $S[X]=S[T(X)]$ for every section $X$. 
The integral that defines the action can be written in coordinates. Thus the action may be seen as a functional $S_{\mathbb{R}^4}$ over sections of a bundle over $\mathbb{R}^4$ instead of the original over $M$. Then the trivial statement that a coordinate transformation $\phi:\mathbb{R}^4\rightarrow \mathbb{R}^4$ induces another action $S'_{\mathbb{R}^4}$ (a different functional) given by $S'_{\mathbb{R}^4}[X]=S_{\mathbb{R}^4}[X\circ \phi]$ is not equivalent to the invariance of the action under the transformation $\phi$ (it would be so only if $S'=S$). Every coordinate transformation induces a new action, but not all of them leave the action invariant (only the Poincaré transformations do).
A: A particle shall have an identity that is kept invariant under symmetry transformations. Irreducible representations provide such an identity, since if you transform a state of a particle you will end up in a state in the same space. On the contrary, if we had not defined particles as irreducible representations, you could apply a symmetry transformation and alter the identity of a particle. You could obtain, for example, a neutrino state by applying a Poincare transformation on an electron state, which is nonsense. Since irreducible representations are indexed by the eigenvalues of the Casimir invariants, the possible values of those invariants index the possible identities of particles. Poincarre symmetry inherits particles with mass and spin, which partially define the identity of a particle and correspond to the two Casimirs of the Poincare group (squared translation generator and squared Pauli-Lubanski vector). The rest of the particle’s identity is given by the Casimirs of other, internal, symmetries. U(1) will give charge, SU(2) isospin and so forth.
