Understanding the definition of a function $\overline f := g'\ ^ T f\circ g$ I have open sets $A,B \in \mathbb R^n$, $g\in C^1(A,B)$,  $g^{-1} \in C^1(B,A)$ and $f: B\rightarrow \mathbb R^n$ is a vector field. 
Now, for a newly defined vector field $\overline f$ with $$\overline f := g'\ ^ T f\circ g$$
I am having troubles understanding what this function actually does. Note that $g ' \ ^T$ denotes the tranpose of the Jacobian of $g$ here. I know that $f\circ g$ maps a vector from $A$ to $\mathbb R^n$ , but how do I multiply a matrix with a function, say: What would $\overline f (t)$ be for a vector $t$?
Any help is greatly appreciated!
 A: For the simplicity of this explanation, suppose $A=B=\mathbb{R}^n$. As you said, $f\circ g$ is a function from $\mathbb{R}^n$ to $\mathbb{R}^n$. Further, $g$ is a function from $\mathbb{R}^n$ to $\mathbb{R}^n$. Given $v\in \mathbb{R}^n$, $g'(v)$ is essentially the best linear approximation (encoded as a matrix) for $g$ at $v$. $g'$ is a map from $\mathbb{R}^n$ to $M_{n\times n}(\mathbb{R})$, and so given a vector $v$, $g'(v)$ is an $n\times n$ matrix with real entries. Since matrices are just lists of numbers which encode information about a linear transformation, put another way, $g'(v)$ is a linear function from $\mathbb{R}^n$ to $\mathbb{R}^n$ (note that we get different such functions for different values of $v$). 
Now we should examine what $(g')^T$ does. Given a linear operator $A$ on a vector space $V$, $A^T$ is a map from the dual space of $V$ (denoted as $V^*$) to itself. Elements of the dual space are functions which take in a vector and give back a real number. So, at a particular $v\in \mathbb{R}^n$, $(g')^T$ takes in a dual vector from $(\mathbb{R}^n)^*$ and gives back another such dual vector. We actually don't need to worry about this technicality, though. So, we're justified in considering $(g')^T(v)$ as a function from $\mathbb{R}^n$ to $\mathbb{R}^n$.
What's not obvious in the notation is that 
$$\bar f(v) = (g')^T(v) \cdot (f\circ g)(v)$$
where the $\cdot$ is effectively matrix multiplication. This should make sense, since $(f\circ g)(v)$ is a vector, $(g')^T(v)$ is a map which takes in vectors and outputs vectors. So, in effect, the RHS takes in a vector and outputs a vector. Thinking of this in terms of matrix multiplication also makes sense; $(g')^T(v)$ is an $n\times n$ matrix, $(f\circ g)(v)$ is a vector, and so the whole RHS outputs a vector. 
