That's probably (no pun intended) because it's not a probability density at all. Wave functions by themselves don't have any physical interpretation. Only their squares do.
The probability of transition is computed like this $$P_{1 \to 2} = {\left \Vert \left < \psi_1 | \psi_2 \right > \right \Vert^2 \over \left \Vert \psi_1 \right \Vert^2 \left \Vert \psi_2 \right \Vert^2}$$
What you've written is just a density of the probability amplitude $\left<\psi_1|x\right>\left<x|\psi_2\right>$. It can be complex so it certainly cannot be probability in any sense.
Edit regarding your additional comments: assuming that $\left \Vert \psi_1 \right \Vert^2 = \left \Vert \psi_2 \right \Vert^2 = 1$ you could write $P_{1 \to 2}$ like this $$P_{1 \to 2} = {\rm Tr} \left( \left|\psi_1\left> \right<\psi_1\right| \left|\psi_2\left> \right<\psi_2\right| \right) = \int \psi_1(y) \psi_1^*(x) \psi_2(x) \psi_2^*(y) {\rm d} x {\rm d} y$$ interpreting it as a trace of the product of projection operators. But this integral representation is basically never done, it's much easier to only compute the amplitude and square it when you're finished.
As a moral, you shouldn't insist on restricting your attention to some portion of position space. Quantum theory "doesn't like it" (except for few special constructions).