Consider the time evolution operator $U(t_f, t_i)$ that controls the evolution of a wave function according to $|\psi(t_f \rangle = U(t_f, t_i) | \psi(t_i) \rangle$.
As I understand it, the Born rule says that we interpret $ \langle x | \psi(t) \rangle$ as the complex probability amplitude of measuring the system $\psi$ to have position $x$ and time $t$. Using the concept of the wave function, we write $\psi(x, t) = \langle x | \psi(t) \rangle$, and is the square norm of this wave function gives us a pdf.
Thus, it seems reasonable to interpret $ \langle x_f | U(t_f, t_i) | \psi(t_i) \rangle$ as the probability amplitude of measuring the postion of $\psi$ to be $x_f$ at time $t_f$. Let us say that $| \psi(t_i) \rangle = | x_i \rangle$. Then it seems reasonable to interpret $ K(x_f, t_f; x_i; t_i) := \langle x_f | U(t_f, t_i) | x_i \rangle$ as the "transition amplitude" of measuring a system to have position $x_f$ at time $t_f$ when we know the system had position $x_i$ at time $t_i$.
However, I have only seen the term "transition amplitude" used (in Townsend's "A Modern Approach to Quantum Mechanics, Chapter 8, and these notes, Section 1, http://www.blau.itp.unibe.ch/lecturesPI.pdf) used to refer to the notation $\langle x_f, t_f | x_i, t_i \rangle$, where the state $ |x_i, t_i \rangle$ is expressly NOT the time evolution of $ |x_i \rangle$ but rather the eigenfunction of the time evolved position operator in the Heisenberg picture (thus, $ |x_i, t_i \rangle = \exp\{i \hat{H} t_i\} | x_i \rangle$).
So, my question (finally) is: why does it seem that the term "transition amplitude" is applied to $\langle x_f, t_f | x_i, t_i \rangle$ and NOT (the equivalent) $K(x_f, t_f; x_i, t_i)$, when it seems that $K(x_f, t_f; x_i, t_i)$ has a very clear understanding as the transition amplitude of a particle starting at position $x_i$ at time $t_i$ to be measured at position $x_f$ at time $t_f$, whereas the interpretation of $\langle x_f, t_f | x_i, t_i \rangle$ seems much less straightforward?
