Confusion with time ordering I am thinking about Proof of correlation function formula in quantum field theory and have realized there is a deeper confusion underpinning that. Consider:
$$T\{U_I(T, t_2)\Phi_I(x_1)\}$$
where $t_2>t_1$ where these are a field operator and the time evolution operator, respectively, both in the Interaction picture.$A=U_I(T, t_2)=T\big\{exp\big(-i\int_{t_2}^{T} dt'\, H_I(t')\big)\big\} = e^{iH_0 (T-t_0)}e^{iH (T-t_2)}e^{iH_0 (t_2-t_0)}$ See here for derivation. 
Now, the first representation with an integral in the exponent, only has operators (in interaction picture) with time in the interval $(T, t_2) > t_1$, so we find:
\begin{align*}
T\{U_I(T, t_2)\Phi_I(x_1)\} &=T\{T\big\{\exp\big(-i\int_{t_2}^{T} dt'\, H_I(t')\big)\big\}\Phi_I(x_1)\} \\
  &=T\big\{\exp\big(-i\int_{t_2}^{T} dt' H_I(t')\big)\big\}\Phi_I(x_1)=U_I(T, t_2)\Phi_I(x_1)
\end{align*}
However, the second representation has only operators in the Schrodinger picture, which are equal to those operators in the interaction picture, at the arbitrary time $t_0$, which may well be chosen to be less than $t_1$, and so we get:
\begin{align*}
  T\{U_I(T, t_2)\Phi_I(x_1)\} &= T\{e^{i(T-t_0)\cdot H_{I,0}(t_0)}e^{i(T-t_2)\cdot H_I(t_0)}e^{i(t_2-t_0)\cdot H_{I,0}(t_0)}\Phi_I(x_1)\} \\
 &=\Phi_I(x_1)e^{i(T-t_0)\cdot H_{I,0}(t_0)}e^{i(T-t_2)\cdot H_I(t_0)}e^{i(t_2-t_0)\cdot H_{I,0}(t_0)} \\
 &=\Phi_I(x_1)U_I(T, t_2) \neq U_I(T, t_2)\Phi_I(x_1)
\end{align*}
 A: The first result is correct. Time ordering acts on interaction picture field operators. $H_I(t)$ is to be understood as a function of interaction picture field operators at time $t$.
The representation
$$ U(t, t') = e^{iH_0t}e^{-iH(t - t')}e^{-iH_0t} $$
does not represent $U$ in terms of interaction picture field operators and is therefore not permissible. Only because the operators are the same at that instant does not mean you can use them as interaction picture operators. The point here is, that the time-marks occur also explicitly, while the time dependence has to be implicit as the time dependence of the field operators for the technique to work.
To correctly evaluate expressions involving time ordering one has to expand the exponential and time order every term:
$$ T\left\{\exp \left(\int_{t_1}^{t_2}dt\,H(t) \right) \phi(t_3) \right\} = \sum_{n=0}^\infty \frac{(-i)^n}{n!} \int_{t_1}^{t_2} dt'_1 \cdots \int_{t_1}^{t_2} dt'_n T H(t'_1) \cdots H(t'_n)\phi(t_3).$$
If now $t_3 < t_1 < t_3$ the operator $\phi(t_3)$ will always be ordered to the right, and therefore can be factored out to the right and so the result is the time evolution operator followed by $\phi(t_3)$.
