Time ordering in correlation function in QFT

In Peskin & Schroeder, "An Introduction to Quantum Field theory", chapter 4, the author derives the 2 point correlation function:

$$\langle \Omega|P{\phi(x)\phi(y)}|\Omega \rangle = \lim_{T\rightarrow \infty(1-i\epsilon)} \frac{\langle 0|P \left [U(T,x^0)\phi_I(x)U(x^0,y^0)\phi_I(y)U(y^0,-T)\right ]|0\rangle}{\langle 0 | U(T,-T)|0\rangle }, \tag{4.30c}$$

where $$P$$ is the time ordering operator and $$U(t,t')$$ is the evolution operator $$U(t,t') =\exp\left \{ -i \int^t_{t'}dt H_I(t)\right \}\tag{4.23}$$ $$\langle \Omega|P{\phi(x)\phi(y)}|\Omega \rangle= \lim_{T\rightarrow \infty(1-i\epsilon)} \frac{\langle 0|P \left [\phi_I(x)\phi_I(y)\exp\left \{ -i \int_{-T}^Tdt H_I(t)\right \}\right ]|0\rangle}{\langle 0 | \exp\left \{ -i \int_{-T}^Tdt H_I(t)\right \}|0\rangle }. \tag{4.31}$$

My question is how do the various $$U()$$ factors like $$U(T,x^0)$$, $$U(x^0,y^0)$$ and $$U(y^0,T)$$ simplify to a final $$\exp\left \{ -i \int_{-T}^Tdt H_I(t)\right \}$$ in the numerator from Eq. (4.30c) to Eq. (4.31)?

My problem is that $$U()$$ does not commute with $$\phi_I$$, so, for example, one cannot really shift $$U(T,x^0)$$ to the right of $$\phi_I(x)$$ in Eq. (4.30c). $$x^0$$ is the $$0^{th}$$ component of $$x$$, and hence $$U(T,x^0)$$ and $$\phi_I(x)$$ are at the same time $$x^0$$, and thus the time ordering operator $$P$$ cannot be used to shift $$U(T,x^0)$$ to the right of $$\phi_I(x)$$. To my understanding, time ordering can be used to shiit terms only if they are at different times (according to their time sequence).

The order you place the operators inside of the time ordering operator doesn't matter. For example $$P[\mathcal{O}(t+1)\mathcal{O}(t)]=P[\mathcal{O}(t)\mathcal{O}(t+1)]=\mathcal{O}(t+1)\mathcal{O}(t).$$ When we write $$P\left [\phi_I(x)\phi_I(y)\exp\left \{ -i \int_T^Tdt H_I(t)\right \}\right ],$$ the time ordering operator rearranges the terms to give $$U(T,x^0)\phi_I(x)U(x^0,y^0)\phi_I(y)U(y^0,-T).$$ Using the former expression is just shorthand.