Remember you have to worry about the temporal inversion by \(\theta\) now. I think it is easiest to see this by explicitly writing it out as seen on page 87.

\[(\theta f)_n (x_1,...,x_n) = f_n(\theta x_1,...,\theta x_n)\] where \(\theta x = (-x^0 , \vec{x})\)

now remember by definition we have \(f_{(a,\mathbf{R})}(x_1,...,x_n)=
f(\mathbf{R}x_1 + a,...,\mathbf{R}x_n + a)\) and (E1) states that the Euclidean Green function is invariant under SO(4) rotations and translations (as stated in your previous question).

Using your notation we have

\((f,\hat{T}^tg) = \sum \sigma_{n+m} (\Theta f^*)_n \times (\hat{T}^t g)_m) = \sum \sigma_{n+m}(\Theta f^*_n \times (g_{(-t,1)})_m)\)

Notice the minus sign on the t, for that is how they define the map, \((\hat{T}^t f)_n (x_1,...,x_n) = f_n(x_1 - t,...,x_n - t)\) where \(t = (t,\vec{0})\)

Now explicitly writing out your expression above we have

\[= \sum \sigma_{n+m}(f^*_n(\theta x_1,...,\theta x_n) \times g_m(x_1 - t,..., x_n-t))\]

We then shift the whole thing by a factor of t, giving us

\[= \sum \sigma_{n+m}(f^*_n(\theta x_1+t,...,\theta x_n+t) \times g_m(x_1 ,..., x_n))\]

\[= \sum \sigma_{n+m}(f^*_n(\theta x_1- \theta t,...,\theta x_n-\theta t) \times g_m(x_1 ,..., x_n))\]

\[= \sum \sigma_{n+m}(\theta f^*_n( x_1- t,..., x_n- t) \times g_m(x_1 ,..., x_n))\]

\[= \sum \sigma_{n+m}(\theta (\hat{T}^t f)^*_n( x_1,..., x_n) \times g_m(x_1 ,..., x_n))\]

Where in the second line we used the fact that t is a purely temporal vector. This gives the result you are looking for in equation (4.7)