My textbook presents the chain rule in case $X \subseteq \mathbb K$ as follows:
In this case, $(g \circ f)^{\prime}(a)=g^{\prime}(f(a)) f^{\prime}(a)$. Clearly, $g^{\prime}(f(a)) f^{\prime}(a)$ denotes the arithmetic product of $g^{\prime}(f(a))$ and $f^{\prime}(a)$. Hence $(g \circ f)^{\prime}(a) = g^{\prime}(f(a)) \cdot f^{\prime}(a)$. It seems to me that $f^{\prime}(a)$ is a value in $\mathbb K$, not a bounded linear map in $\mathcal L(\mathbb K, \mathbb K)$. Similarly, $g^{\prime}(f(a))$ is a value in $E$, not a bounded linear map in $\mathcal L(\mathbb K, E)$.
Here is its general form:
In my understanding, $\partial f\left(x_{0}\right) \in \mathcal L(E,F)$ and $\partial g\left(f\left(x_{0}\right)\right) \in \mathcal L(F,G)$. As such, $\partial g\left(f\left(x_{0}\right)\right) \partial f\left(x_{0}\right)$ does not denote the arithmetic product of $\partial g\left(f\left(x_{0}\right)\right)$ and $\partial f\left(x_{0}\right)$. Instead, it is the function composition of $\partial g\left(f\left(x_{0}\right)\right)$ and $\partial f\left(x_{0}\right)$, i.e. $$\partial g\left(f\left(x_{0}\right)\right) \partial f\left(x_{0}\right) = \partial g\left(f\left(x_{0}\right)\right) \circ \partial f\left(x_{0}\right) \in \mathcal L(E,G)$$
My question:
Could you please elaborate on how the chain rule in these two cases coincide? Thank you so much for your help!