I'm trying to understand the proof of Lemma 6 (p. 27) in "An Introduction to Rough Paths" - A. Lejay.
He first defines the $\mathbf{y}_{s, t}^1$ to be $$\mathbf{y}_{s, t}^1 = f(x_s) \mathbf x_{s, t}^1 + \nabla f(x_s) \mathbf x_{s, t}^2,$$ where $\mathbf x_{s, t}^1$ are the path increments and $\mathbf x_{s, t}^2$ are the (postulated) first iterated integrals. In the beginning of the proof of Lemma 6, he then says $$\mathbf{y}_{s, u}^1 \otimes \mathbf{y}_{u, t}^1 = f(x_s) \mathbf x_{s, u}^1 \otimes f(x_u) \mathbf x_{u, t}^1.$$
Now, my question is what happens to the other terms? I understand, that in the truncated tensor algebra $T_2(\mathbb R^n)$, all terms of tensor level $\geq 3$ are ignored, but for that to be used, one would have to pull the $\nabla f(x_s)$ out of the tensor product and I don't see why this should be allowed.