Why is this combination of a covariant derivative and vector field a (1,1)-tensor? I have a question regarding something Penrose says in section 14.3 of The Road to Reality.
It says '...when $\nabla$ acts on a vector field $\xi$, the resulting quantity $\nabla \xi$ is a $(1,1)$-valent tensor.'
I understand that $\nabla$ is a $(0,1)$-tensor and $\xi$ as a vector field is a $(1,0)$-tensor, so it sort of makes sense that $\nabla \xi$ is a $(1,1)$-tensor.  However, what's throwing me off is the wording "acts on".  I understand that the idea of $\nabla$ is that it gives a notion of derivative not just for scalar fields, but vector fields and general tensors.  It seems "acts on" implies that $\nabla \xi$ should be a $0$-tensor.
So my question is: is this just a poor choice of language, and is $\nabla \xi$ just $\nabla$ with respect to a vector field (instead of just a single vector) $\xi$, or is the actual evaluation of $\xi$ by $\nabla$ not a scalar field as I would expect, but actually a $(1,1)$-tensor?
Thanks.
EDIT: I have received an answer to my question (see comment below), but now I'd really like more detail behind how, if I may ask in this same question, the $(1,1)$-tensor $\nabla \xi$ describes how $\xi$ changes from point to point (intuitively and concretely, preferably, but at least intuitively).

A $(1,1)$-tensor can be thought of as a linear map that sends vectors to vectors; so given a vector $X$ based at $p$, $\nabla\xi(X)=\nabla_X \xi$ will be another vector based at $p$, which you should think of as the change in the vector field $\xi$ when you move a small amount in the direction $X$ starting from $p$.
