Does the quadratic mass term $\frac{1}{2}m^2\phi^2$ belong to the free Lagrangian or is it an interaction term?

Conventionally, bilinear terms in the Lagrangian are called kinetic terms; the rest (terms with three or more fields) is considered interactions. The reason is simple: if you solve the theory with just the kinetic terms, the solution describes the behavior of free (non-interacting) particles. For the example you provided, the solution is that of a free scalar particle and the $m$ in front of the $\phi^2$ term happens to be the mass of the particle. That is why the term is called a mass term, part of the kinetic terms.

Nonetheless, it is just a convention. If you insist, you can treat the term as a self-interaction. But then when you start solving the theory, you would discover that the self-interaction term give the particle, which you thought was massless, a mass.

You definitely can treat it as an interaction. If you do, the "free" propagator is that of a massless theory: $$G_0(k^2) = {1 \over k^2 + i \epsilon}$$

The "interacting" propagator can be summed geometrically:

$$G_i(k^2) = G_0 + G_0 {m^2} G_0 + ...$$ $$ G_i(k^2) = {G_0 \over 1 - G_0 m^2} = {1 \over k^2 - m^2 + i \epsilon}$$

However the main reason why we treat the "free" propagator as anything quadratic in the field is because anything that is bilinear in the fields can be evaluated exactly. It is just a gaussian integral:

$$ Z(J) = \int \mathcal{D} \phi e^{i \phi M \phi + J \phi}$$ $$ Z(J) \propto {1 \over \sqrt{\det M}} e^{\frac12 J M^{-1} J}$$

Because it can be solved exactly, we consider it "free" solution, from which we compute perturbatively interactions.

Because the solutions are diagonalized in k-space, the propagation of the eigenstates do not scatter each other.