Let $e^{-it\hat{H}/\hbar}$ be the time evolution operator for a Hamiltonian $\hat{H}$ and $K(x,t)$ its associated integral kernel, i.e. $$\varphi(x,t) = e^{-it\hat{H}/\hbar}\varphi_0(x) = \int_{\mathbb{R}} K(x-y, t)\varphi_0(y) dy. \tag{1}$$
Since $K(x,t)$ is a fundamental solution (Green's function) to the Schrodinger equation, the time evolution of a state $\psi$ can be obtained by convolution with $K(x,t)$ as in (1). The important point here is that $K$ being written as $K(x-y, t)$ is a consequence of convolution.
In many physics textbooks they take this a step further and give physical meaning to $K(x-y, t)$. They state it is the probability of a particle going from $x$ to $y$ in time $t$. For example, in Sakurai's textbook he gives the formula $$K(x'', t; x', t_0) = \langle x'', t| x', t_0 \rangle$$ where $$K(x'', t; x', t_0) = \sum_a' \langle x'' | a'\rangle \langle a' | x'\rangle e^{iE_{a'}(t-t_0)/\hbar}.$$
In my notation Sakurai's result is: $$K(x'', t; x', t_0) = K(x''-x', t-t_0) = \langle x'', t| x', t_0 \rangle.$$
I do not see what justifies $K(x-y, t)$, which originally appeared as a consequence of convolution, can be interpreted as a probability for a particle to go from $x$ to $y$ (after all the $y$ is only an integration variable that could've been chosen to be anything). I come from a background in mathematics and not physics so I find Sakurai's explanation very hard to follow.
