How should I be thinking about the total derivative at a point? I am curious as to how I should be reasoning about the total derivative of a function when evaluated at a point. I have been thinking of these objects as linear functions, which it seems to me that they are. However I just happened across the following problem which suggests that I should instead be thinking of them in the same way I think of matrices:
Say I have two functions $f$ and $g$ such that 
$$g, f \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$$ 
Now, if $\chi$ is a point in $\mathbb{R}^n$ we see that
$$Dg(\chi), Df(\chi) \colon \mathbb{R}^n \rightarrow \mathbb{R}^m$$
are linear transformations. Now, if $v$ is another vector in $\mathbb{R}^n$ I wonder how I should interpret the expression
$$[Dg(\chi)\cdot Df(\chi)](v)$$
For, if I think of $Dg(\chi)$ and $Df(\chi)$ as matrices, we have that
$$[Dg(\chi)\cdot Df(\chi)](v) = Dg(\chi)(Df(\chi)(v))$$
(ie a composition of functions). If I instead regard them as linear functions then I should have
$$[Dg(\chi)\cdot Df(\chi)](v) = Dg(\chi)(v) \cdot Df(\chi)(v)$$
which doesn't make sense.
It seems to me that the notation is vague. Of course it may just be that I am being dense. How should I reason with these derivatives? 
I apologise if the question is somewhat vague. Feel free to add the (soft-question) if it is necessary.
 A: 
I wonder how I should interpret the expression $[Dg(\chi)\cdot Df(\chi)](v)$

I think it's best to not try to interpret it at all. It's a  unnatural  expression without an intrinsic meaning. 
Multiplication of matrices is composition of linear operators. The linear operators $Df(\chi)$   and $Dg(\chi)$ are both defined in the same copy of $\mathbb R^n$ (the domain of $f$ and $g$) and go from there to the second copy of $\mathbb R^n$, in which $f$ and $g$ take values. Their composition does not make real sense (although it's formally defined as a product of matrices). 
On the other hand, if we are in the situation of the chain rule - the domain of $g$ contains the range of $f$ - then the product of $Dg(f(\chi))$ with $Df(\chi)$  makes sense: it's the composition of linear operators.  We have three copies of $\mathbb R^n$ in this situation: $f$ goes from first to second, $g$ from second to third.
It is important to keep in mind the distinction between different copies of $\mathbb R^n$. Doing  calculus on manifolds enforces this distinction: if $f$ and $g$ are smooth maps from manifold $M$ to manifold $N$, there is no way to multiply $Df$ and $Dg$. 

On the third hand, if we are talking about maps between Riemannian manifolds (which include $\mathbb R^n$ with their standard metric), then one can take the scalar product of $Df(\chi)$ and $Dg(\chi)$. It is usually defined as $\langle A,B \rangle =\operatorname{tr}( B^TA)$ where $A$ and $B$ are matrices (such as $Df(\chi)$ and $Dg(\chi)$). This product generalizes $f'(x)g'(x)$ from one-dimensional analysis.
