For vector space $\mathbb{R}^n$ we have partial derivatives, which obey the chain rule, e.g:
let $F:\mathbb{R}^n \to \mathbb{R}^m$, $f:\mathbb{R}^m\to \mathbb{R}$, assume standard basis for $\mathbb{R}^n$ is $x^i$ and standard basis for $\mathbb{R}^m$ is $y^j$.So for composition we have:
$$\left.\frac{\partial}{\partial x^{i}}\right|_{p}(f \circ F)=\frac{\partial f}{\partial y^{j}}(F(p)) \frac{\partial F^{j}}{\partial x^{i}}(p)$$
which is the standard chain rule.
Now consider the general case derivative as linear map between algebra $v:A\to B$ with $v(fg) = fv(g)+gv(f)$.
In this case does chain rule for composition $v(f\circ g)$ still hold? It seems not?
(we know for differential $dF_p:T_pM\to T_p N$ chain rule still holds)
Best Answer
In the case of smooth manifolds, what you call the chain rule is a manifestation of functoriality of the functor taking a manifold with marked point $(M,p)$ to its tangent space $T_pM$ and taking a smooth map of such objects $f:(M,p)\to (N,q)$ to the associated differential $df_p:T_pM\to T_qN$. Functoriality says that given a composition $$ (M,p)\xrightarrow{f} (N,q)\xrightarrow{g}(P,r)$$ there is a relation $d(g\circ f)_p=dg_q\circ df_p$. In less abstruse language, this just says that the differential of the composition is the composition of the differentials. Putting that into concrete terms, given $$ \Bbb{R}^n\xrightarrow{F} \Bbb{R}^m\xrightarrow{f} \Bbb{R}$$ as above, we know that the differentials are respectively $$ \bigg[\frac{\partial F^i}{\partial x^j}\bigg]_p$$ and $$ \bigg[\frac{\partial f}{\partial y^i}\bigg]_{F(p)}$$ where the coordinates on the first space are $x^1,\ldots, x^n$ and the coordinates on the second space are $y^1,\ldots, y^m$ and the first matrix is $m\times n$, and the second is $1\times m$. The composite of the differential is the multiplication of these matrices, which is as you write $$ \bigg[ \sum_{i=1}^n\frac{\partial F^i}{\partial x^j}(p)\frac{\partial f}{\partial y^i}(F(p))\bigg]$$ where this is an $1\times n$ matrix.
The question you are asking is different. Let's say that $A$ and $B$ are $k-$algebras for some field $k$. Then a morphism $v:A\to B$ which is $k-$linear and Leibniz (i.e. $v(fg)=v(f)g+fv(g)$) is a type of differential operator. However, here it is unclear what you want the chain rule to mean. The chain rule is what occurs when we apply a differential operator to a composite of functions in our manifold setting. In this case, $f\circ g$ does not even make sense a priori.
I make the following proposal: Given a category of geometric spaces $\mathscr{C}$, and a "function" $F: \mathscr{C}\to \mathscr{A}$, assigning to each space $X$ an algebraic structure $F(X)$, we say that $F$ obeys a chain rule if $F$ is functorial in the sense of above: given $$ X\xrightarrow{f}Y\xrightarrow{g}Z$$ we have $F(g\circ f)=F(g)\circ F(f)$. This is admittedly a bit vague, but it illustrates what we "used" to define the chain rule.