Two different interpretations of Feynman's path integral?

In quantum mechanics, we have Feynman’s path integral:
begin{eqnarray}
langle x_{f},t_{f}| x_{i},t_{i}rangle = int {cal D}x expbigg{(}iint_{t_{i}}^{t_{f}}dt, L(x,dot{x}) bigg{)}tag{1}label{1}
end{eqnarray}
Here, the notation means:
$$|x,trangle = e^{frac{i}{hbar}tH} |xrangle$$
in Heisenberg’s picture. However, in my class notes, my professor proved that:
begin{eqnarray}
U_{t}(x,y)= int {cal D}x expbigg{(}iint_{t_{i}}^{t_{f}}dt,L(x,dot{x}) bigg{)}tag{2}label{2}
end{eqnarray}
where $U_{t}(x,y)$ is the integral kernel of $e^{frac{i}{hbar}tH}$, viewed as an operator on $L^{2}(mathbb{R}^{d})$.

My problem is: $langle x_{f},t_{f}|x_{i},t_{i}rangle$ and $U_{t}(x,y)$ seem to be two different objects, although the path integral representation is the same. What am I missing here? Are both these objects the same?