As @ArnoldNeumaier stated you need to use the definition of (4.3) and (E1). Where (4.3) is

\[(f,g) = \sum_{n,m} G_{n+m}(\theta f_n^* \times g_m)\]

and (E1) states that the Euclidean Green function is invariant under SO(4) rotations and translations. \(G_n(f) = G_n(f_{(\underline{a}, \mathbf{R})})\) where \(f_{(\underline{a}, \mathbf{R})}(\underline{x_1},...,\underline{x_n})= f(\mathbf{R}\underline{x_1} + \underline a,...,\mathbf{R}\underline{x_n} + \underline a)\).

While \(\hat U_s(\vec a)\) generates purely spatial translations on the functions it acts on. Now we can see how the statement follows.

\[(f,\hat U_s(\vec a)g) = \sum_{n,m} G_{n+m}(\theta f_n^* \times (\hat U_s(\vec a)g)_m) = \sum_{n,m} G_{n+m}(\theta f_n^* \times g_{(-\vec {a}, 1)m}) \]

Then we use (E1) to shift the whole Green function by a value \(\vec a\), where we don't need to worry about \(\theta\) for it is just temporal inversion, which commutes with spatial shifts. Thus we find

\[\sum_{n,m} G_{n+m}(\theta f_n^* \times g_{(-\vec a, 1)m}) = \sum_{n,m} G_{n+m}(\theta f_{(\vec a, 1)n}^* \times g_{m})= \]

\(= \sum_{n,m} G_{n+m}(\theta (\hat U_s(-\vec a)f_{n})^* \times g_{m}) =(\hat U_s(-\vec a)f , g)\)