Consider a system whose state is initially $\left|\psi(t_i)\right\rangle$. At a later time $t_f$, its state will be $$\left|\psi(t)\right\rangle=\mathcal{U}(t_i,t_f)\left|\psi(t_i)\right\rangle$$ where $\mathcal{U}(t_i,t_f)=\exp(-i\hat{H}(t_f-t_i)/\hbar)$ is the time evolution operator (in the case of a time-independent Hamiltonian).
Now consider the matrix elements of $\mathcal{U}(t_i,t_f)$ in the position eigenbasis $\{\left|x\right\rangle\}$. We define the propagator as $$\left\langle x_f\middle|\mathcal{U}(t_i,t_f)\middle|x_i\right\rangle$$
How can I show that the propagator as defined above can also be written as the transition amplitude $$\left\langle x_f,t_f\middle|x_i,t_i\right\rangle$$ ?
Update
After thinking about it for a bit, I came up with this: since
$$\left|x_i,t_i\right\rangle=\exp(-i\hat{H}t_i/\hbar)\left|x_i\right\rangle$$
and
$$\left|x_f,t_f\right\rangle=\exp(-i\hat{H}t_f/\hbar)\left|x_f\right\rangle\quad\Rightarrow\quad\left\langle x_f,t_f\right|=\left\langle x_f\right|\exp(i\hat{H}t_f/\hbar)$$
then it follows that $$\left\langle x_f,t_f\middle|x_i,t_i\right\rangle=\left\langle x_f\middle| \exp(-i\hat{H}(t_i-t_f)/\hbar) \middle| x_i\right\rangle$$
however this produces the wrong sign in the exponent, and I also believe I might be mixing the Heisenberg and Schrodinger pictures.