Q:

Why is mass the quadratic term in a Lagrangian? 

Why is mass the quadratic term in a Lagrangian? 

A:

I'm not sure if this is the sort of answer you want, but here goes. 
Mass, by definition, is the thing in $F=ma$. That is, it's the coefficient in front of the acceleration in the equation of motion.  If you build a Lagrangian out of terms involving a coordinate $x$ and its derivative $\dot x$, and then work out the Euler-Lagrange equations, the only way to get a term of the form $m\ddot x$ at the end is to start with a term of the form ${1\over 2}m\dot x^2$ in the Lagrangian.  
That's the classical-mechanics answer.  If you were looking for a field-theory answer instead, the details are different, but the idea is the same.  Think about how you expect mass to show up in the field equation, and work out what term in the original Lagrangian will give rise to the appropriate expression when you use the Lagrangian to derive the field equation. 

A:

For a relativistic particle (free or in an external filed), the Lagrangian contains mass differently, not in a "quadratic" way. 

A:

Take the Lagrangian with only the kinetic and quadratic terms.  Find the wave equation and propagator and you'll see that they correspond to those of a free relativistic particle with mass squared equal to the quadratic coefficient.  So the interpretation is correct in free field theory. 
Adding higher order terms doesn't change this interpretation, since the higher order terms correspond to interactions.