What does it mean by spin 1/2 or spin 2 field? I see in common discussions people simply use the terminology spin 1/2 field or spin 2 fields as if it is some common term like hamiltonian. How to think about these fields and understand what it means when we say gravitation is a spin 2 field.
 A: Relativistic fields are classified by how they transform under the Lorentz group. This means they are classified by representations of the universal cover of the Lorentz group ${\rm SL}(2,\mathbb{C})$. The irreducible ones are classified by two numbers $A,B\in \frac{1}{2}\mathbb{Z}_+$ and denoted $(A,B)$.
Since the Lorentz group contains the rotation group, the universal cover of the Lorentz group contains the universal cover of the rotation group, ${\rm SU}(2)$. In particular, this means that when we specify a field, we are also saying how it transforms under rotations. But from the theory of angular momentum studied in non-relativistic QM we know that the irreducible representations of ${\rm SU}(2)$ are the various spin $j$ representations, where $j\in \frac{1}{2}\mathbb{Z}_+$.
Now, pick one particular relativistic field. It takes values in a certain vector space $V$ which carries a representation of the Lorentz group. By the above argument it carries a representation of ${\rm SU}(2)$ and therefore it decomposes under ${\rm SU}(2)$ $$V=\bigoplus_{j} V_j$$
for some set of $j$'s. This is relevant because in relativistic quantum field theory the $j$'s appearing above are the possible spins of particles such a field may encode.
Examples:

*

*The scalar field. The relevant representation is the trivial scalar representation. It has only $j=0$ and only encodes particles without spin.


*The Weyl spinors. The relevant representations are the irreducible $(\frac{1}{2},0)$ and $(0,\frac{1}{2})$ representations. One finds that the only possible $j$ is $j=\frac{1}{2}$. This is a spin $\frac{1}{2}$ field and is capable of encoding spin $\frac{1}{2}$ particles.


*The vector field. The relevant representation is $(\frac{1}{2},\frac{1}{2})$. One finds that the possible $j$'s here are $j=0,1$. This field is capable of encoding both spin $0$ and spin $1$ particles. It is easy to understand the spin $0$ appearing here. If you have a scalar field $\phi(x)$ then $\partial_\mu \phi(x)$ is a vector field. Nevertheless, since $\phi(x)$ only encoded spin zero particles, the same will be true of $\partial_\mu \phi(x)$.


*The $(A,B)$ fields. The generalization of the above are the $(A,B)$ representations already mentioned. One may show that the possible $j$'s are $j=A+B,\dots, |A-B|$. It is no coincidence that this resembles the sum of angular momenta. The mathematical structure behind is indeed the same.
When a field in a certain representation is constructed exactly to encode a particle of a spin $j$ which is allowed by the above analysis we call it a spin $j$ field. That is I believe the most to the point answer to the question.
For more details I strongly recommend Weinberg's The Quantum Theory of Fields, Chapter 5.
A: In geometrical view, spin $s$ particle returns its original state when it is rotated as $2\pi / s$. For example, spin $1/2$ has minus sign with rotating as $2\pi$, but it returns original state with rotating as $4\pi $. Polarization state of gravitational wave which comes from graviton, spin $2$ particle, returns its original state with rotating as $\pi $.
There is other point view: transformation of tensor and spinor. Spin $2$ particle can be expressed with order of 2 tensor like as $g_{\mu\nu}$. It satisfies transformation law of tensor. For spin $1/2$ case, transformation of spinor can be expressed with some gamma matrices. ($\psi \rightarrow \Lambda_{1/2} \psi$, and $\Lambda_{1/2}$ can be defined with some $\gamma^{\mu}$. See Quantum Field Theory, Peskin & Schroeder, sec 3.2.)
Also, such spin $0, \; 1/2, \; 1, \; 3/2, \; 2, ...$ particles can be interpreted as representation of $so(3)$ Lie algebra. $SO(3)$ Lie group is the collection of unitary transform relevant with 3D rotation, and $so(3)$ Lie algebra is their generators with well-defined commutator relation. We can find some matrix representation of $so(3)$, and find their eigenstates. These states are corresponding to each spin state. (This story is well described in most quantum mechanics book, spin system chapter.)
