When you substitute in the change of variable into the governing equation, you get a (dx/dz)^2 term.   If you define x as a piecewise function then it's derivative isn't simply k1 or k2. 

dx/dz becomes (dk/dz*z+k) from the chain rule, dk/dz is where the recursion happens becuase z cannot be eliminated (z = x/k(z) = x/k(x/k(z)) = etc)

That's correct, Match1 is -k1*D[1](T1)(L1,t) = -k2*D[1](T2)(L1,t).

The change of variable idea is intersting, when I substitute in x=k*z, after some chain rule fun I get the following PDE:

diff(T(x,t),t) = k/rho/Cp * diff(x(z),z)^2 * diff(T(x,t),x$2)

If I assume that diff(x(z),z) = k (that is, k is constant) then I'm all set, but if we want to solve over the entire domain then k = k(z) and is a step function at z = L1.  Trying to write this in terms of k(x) is an issue becuase x is a function of k and it becomes a recursive nightmare.

Perhaps there is a way around this...