Let $E,F$ be two finite-dimensional Banach spaces, with $U\subseteq E$ open and $L(E,F)$ the collection of linear maps $E\to F$. In the following excerpt (from Abraham's Foundations of Mechanics), the authors speak of a function $U\to L(E,F)$ being continuous, implying there is some topology on $L(E,F)$. What is such topology?
Furthermore, why is the definition below equivalent to the usual definition of $f$ being $C^1$ iff its partial derivatives exist and are continuous on $U$ ?
By all standard accounts, we should want $\mathscr{L}(X;Y)$ to be the space of continuous (i.e. bounded) linear maps $X\to Y$ for normed linear spaces $X,Y$. We topologise this by the metric topology; indeed $\mathscr{L}(X;Y)$ is itself a normed linear space with the norm-metric topology. What's the norm? Well, $\|T\|$ for $T\in\mathscr{L}(X;Y)$ is just the operator norm of $T$, $\|T\|:=\sup_{x\in X\\\|x\|\le1}\|T(x)\|$. This exists since $T$ is by definition continuous/bounded.
This is known as the strong topology on $\mathscr{L}(X;Y)$, sometimes. There exist other topologies, e.g. the weak operator topology, but in the context of (Frechet) derivatives we use the strong topology.
If for $X=\Bbb R^n,Y=\Bbb R^m$ and $f:\Bbb R^n\to\Bbb R^m$ you consider $f$ to be $C^1$ iff. all of its partial derivatives exist and are continuous, we can show this is equivalent to $f$ having a continuous derivative $\Bbb R^n\to\mathscr{L}(\Bbb R^n;\Bbb R^m)$.
This is almost certainly a duplicate but a quick search didn't bring anything up- the rest of the details can be found here, I'll just explain why $\mathrm{d}_f$ can be viewed as the matrix of partial derivatives (the Jacobian).Firstly assume $\mathrm{d}_f:\Bbb R^n\to\mathscr{L}(\Bbb R^n;\Bbb R^m)$ exists and is continuous. For any $1\le i\le m,1\le j\le n$ and any $x\in\Bbb R^n$, $\mathrm{d}_f(x)$ is a linear map $\Bbb R^n\to\Bbb R^m$ and we can express it as a $m\times n$ matrix in the canonical basis. Let $a_{i,j}(x)$ be its $(i,j)$th element. We would like to show that this defines a continuous function $a_{i,j}$ and that it serves as $\partial_j^if(x)$.
I can't see your "$1.3.2$" but it's almost certainly something like this: $$\lim_{y\to x}\frac{f(y)-f(x)-\mathrm{d}_f(x)(y-x)}{\|y-x\|}=0$$
In particular, taking $y=x+(0,\cdots,0,h,0,\cdots,0)$ as $h\to 0$, where $h$ lives in the $j$th coordinate, we find: $$\lim_{h\to0}\frac{f(x_1,\cdots,x_j+h,\cdots,x_n)-f(x)-(a_{1,j}(x)\cdot h,\cdots,a_{i,j}(x)\cdot h,\cdots,a_{m,j}(x)\cdot h)}{|h|}=0$$Taking $j$th coordinates tells you exactly that $\partial_i(f_j(x))$ exists and equals $a_{i,j}(x)$ (you can replace $|h|$ with $h$ if the limit equals zero), noting that $|\text{something}|\to0$ iff. $\text{something}\to0$.
It's also a good exercise to check that (the linear operators represented by) $A_k$ tends to $A$ in $\mathscr{L}(\Bbb R^n;\Bbb R^m)$ iff. the entries of $A_k$ tend to the entries of $A$ (as a usual limit of a real sequence). Then continuity of $\mathrm{d}_f$ is equivalent to continuity of the partial derivatives (if $\mathrm{d}_f$ actually exists).