First of all, charts are there because calculus on manifolds is (typically) done by way of focusing on patches small enough that coordinates are well-defined. So let's focus on the flat case.
Let $U\subseteq \mathbb{R}^n$ and $V\subseteq\mathbb{R}^m$ be open subsets, fix $r,s\in \mathbb{Z}_{\geq0}$ with $r\geq s$ and consider the set $C^r(U,V)$ of all $r$-times continuously differentiable functions from $U$ to $V$. We would like to endow $C^r(U,V)$ with a $C^s$-topology of sorts, which should incorporate the first $s$ derivatives when it comes to proximity, the zeroth derivative of a function being the function itself. In other words for two functions $f,g\in C^r(U,V)$, we want
$$f\approx_{C^s} g \iff f \approx g, f' \approx g', f'' \approx g'', ... , f^{(s)}\approx g^{(s)},$$
where $\approx$ means "approximately", and derivatives of functions are certain multilinear functions (e.g. as in Lang's Fundamentals of Differential Geometry).
We could consider the product topology, which would give the topology of pointwise convergence of each derivative. Or we may use the supremum (i.e. $C^0$-) norm for each one of the derivatives, which would give the topology of uniform convergence. In fact since all the functions on the RHS are continuous if we restrict our attention to compact subsets $K\in\mathcal{K}(U)$ of $U$ the supremum (semi-)norm becomes nicer to deal with, and we get a topology more rigid (= stronger = finer = larger = more expensive = higher resolution = harder convergence) than the product topology and more flexible (= weaker = coarser = smaller = cheaper = lower resolution = easier convergence) than the uniform topology. See the discussion here for further details on this comparison.
To that end define a one parameter family $\mathfrak{g}_\bullet^s$ of seminorms on $C^r(U,V)$ by
$$\mathfrak{g}_\bullet^s:\mathcal{K}(U)\to F(C^r(U,V), \mathbb{R}_{\geq0}),\quad
K\mapsto \left[f\mapsto \max_{k\in\{0,1,...,s\}}\max_{x\in K}\left|f^{(k)}(x)\right|\right].$$
(Here the target is the set of all functions from $C^r(U,V)$ to $\mathbb{R}_{\geq0}$.)
Using this family of seminorms parametrized by compact subsets instead of just one norm (e.g. the supremum norm on the whole open set $U$) gives uniform convergence on compacta (or compact convergence), and the topology $\mathcal{N}^s$ generates is precisely the topology of uniform convergence of the first $s$ derivatives on compacta.
Observe that this is not the unique family of seminorms that does the job that we set forth above (though taking the maximum over the order of derivatives has the benefit of playing nicely with intersection). Then the weak subbasic $C^s$-neighborhoods $\mathcal{N}^s$ of Hirsch would precisely be
$$\mathcal{N}^s: C^r(U,V)\times \mathcal{K}(U)\times ]0,\infty]\to \mathcal{P}(C^r(U,V)), \quad (f,K,\varepsilon)\mapsto \{g\in C^r(U,V)\mid \mathfrak{g}_K^s(f-g)<\varepsilon\}.$$
(Here the target is the set of all subsets of $C^r(U,V)$.)
To generalize this to the manifold case we need to replace $C^r(U,V)$ by $C^r(M,N)$ for $M,N$ two manifolds. But the condition for a function to be in $C^r(M,N)$ is it being in $C^r(U,V)$ where $U\in\operatorname{Chart}(M)$ and $V\in\operatorname{Chart}(N)$, which is why the neighborhood-valued map $\mathcal{N}^s$ now gets two additional parameters $\operatorname{Chart}(M)$ and $\operatorname{Chart}(N)$.
Now let's focus on Rudin's definition of weak topology. Let me rephrase it in a more categorical way (seeing spaces parametrized by functions bothers me; the targets and functions come at the same time, the way I see it).
Consider the forgetful functor $\operatorname{Forget}: \operatorname{Top}\to \operatorname{Set} , (S,\mathcal{T}(S))\mapsto S$, let $X\in \operatorname{Set}$ and consider the comma category $(X\searrow \operatorname{Forget})$. An object in $(X\searrow \operatorname{Forget})$ is a pair $(f,Y)$ where $Y$ is a topological space and $f:X\to\operatorname{Forget}(Y)$ is a function.
For any $(f,Y)\in(X\searrow \operatorname{Forget})$, we could pull back the topology of $Y$ via $f$ (i.e. we could look at preimages; which are denoted by $^\ast$ throughout) and we would get a subset $f^\ast(\mathcal{T}(Y))$ of $\mathcal{P}(X)$. Likewise for any collection $\mathfrak{F}\subseteq(X\searrow \operatorname{Forget})$ we can put all the pullbacks together and consider
$$\bigcup_{(f,Y)\in\mathfrak{F}}f^\ast(\mathcal{T}(Y))\subseteq \mathcal{P}(X).$$
Let us denote by $\mathcal{T}(X,\mathfrak{F})$ the smallest topology of $X$ that contains $\bigcup_{(f,Y)\in\mathfrak{F}}f^\ast(\mathcal{T}(Y))$, i.e.
$$\mathcal{T}(X,\mathfrak{F})=\bigcap\left\{\mathcal{T}\subseteq\mathcal{P}(X)\left\vert \mathcal{T}\supseteq \bigcup_{(f,Y)\in\mathfrak{F}}f^\ast(\mathcal{T}(Y))\mbox{ and }\mathcal{T}\mbox{ is a topology}\right\}\right..$$
There are certain universal properties associated to $\mathcal{T}(X,\mathfrak{F})$ as expected. In particular $\mathcal{T}(X,\mathfrak{F})$ is the smallest topology that makes all maps $f:(X,\mathcal{T}(X,\mathfrak{F}))\to (Y, \mathcal{T}(Y))$, $(f,Y)\in\mathfrak{F}$, continuous. See wikipedia for further categorical musings. Also see the discussion here.
Finally let's put these two things together. We are trying to topologize $C^r(U,V)$, so set $X:=C^r(U,V)$. Here are some families in $(C^r(U,V)\searrow \operatorname{Forget})$ that are worthy of mention in our discussion:
- $\mathfrak{F}_1^s:=\{(\mathfrak{g}_{\{x\}}^s,\mathbb{R}_{\geq0})\mid x\in U\}$ (singletons are compact subsets)
- $\mathfrak{F}_2^s:=\{(\mathfrak{g}_K^s,\mathbb{R}_{\geq0})\mid K\in\mathcal{K}(U)\}$ (seminorms we defined above)
- $\mathfrak{F}_3^s:=\{(\mathfrak{h}_{f,K}^s:=\mathfrak{g}_K^s(\bullet-f),\mathbb{R}_{\geq0})\mid K\in\mathcal{K}(U), f\in C^r(U,V)\}$ (translates of the seminorms in $\mathfrak{F}_2^s$)
- $\mathfrak{F}_4^s:=\left\{\left(g\mapsto \max_{k\in\{0,1,...,s\}}\sup_{x\in U}\left|g^{(k)}(x)-f^{(k)}(x)\right|, \mathbb{R}_{\geq0}\right)\mid f\in C^r(U,V)\right\}$
- $\mathfrak{F}_5^s:=\left\{\left(D^k,C^0\left(U,\operatorname{Lin}\left({(\mathbb{R}^n)}^{\otimes k}, \mathbb{R}^m\right)\right)\right)\mid k\in \{0,1,...,s\}\right\}$ ($D^k: f\mapsto f^{(k)}$ is the $k$th derivative operator)
In the first four options the targets are topologized in the expected manner. They are somewhat more general in that their analogs apply to more general situations. The final option is making use of differentiation specifically. Here the set $\operatorname{Lin}\left({(\mathbb{R}^n)}^{\otimes k}, \mathbb{R}^m\right)$ of $k$-linear functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ are topologized by the operator norm (say), and the target set is endowed with the compact-open topology (= $\mathcal{N}^0$ topology = topology of uniform convergence on compacta). I have suppressed some topolinear isomorphisms and embeddings here because I think I'm already pushing your patience.
It is straightforward that $\mathcal{T}(C^r(U,V),\mathfrak{F}_1^s)$ is the topology of pointwise convergence of the first $s$ derivatives and that $\mathcal{T}(C^r(U,V),\mathfrak{F}_4^s)$ is the topology of uniform convergence of the first $s$ derivatives.
Here are some observations to compare the remainder (I'll suppress the quantifiers for the sake of sanity):
Observation 1: $\mathcal{N}^s(f,K,\varepsilon) = (\mathfrak{h}_{f,K}^s)^\ast([0,\varepsilon[).$
Observation 2: $|\mathfrak{h}_{f,K}^s(g)-\mathfrak{h}_{f,K}^s(h)|\leq \mathfrak{g}_K^s(g-h).$
Observation 3: $|\mathfrak{g}_K^s(f)-\mathfrak{g}_K^s(g)|\leq \mathfrak{g}_K^s(f-g).$
Observation 4: $\mathfrak{h}_{0,K}^s=\mathfrak{g}_K^s$, provided that $0\in V$.
Observation 5: $\mathcal{N}^s(f,K,\varepsilon)=\bigcap_{k\in\{0,1,...,s\}}(D^k)^\ast(\mathcal{N}^0(f^{(k)},K,\varepsilon))$.
All these observations are straightforward to verify.
Observation 1 shows that the $\mathcal{N}^s$-topology is contained in any topology w/r/t which the $\mathfrak{h}_{\bullet,\bullet}^s$ are continuous, hence
$$\mathcal{N}^s\mbox{-topology } \subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s).$$
Observation 2 shows that the $\mathfrak{h}_{\bullet,\bullet}^s$ are ($1$-Lipschitz) continuous when the domain is endowed with the $\mathcal{N}^s$-topology, so
$$\mathcal{N}^s\mbox{-topology } \supseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s).$$
Likewise, by Observation 3 the $\mathfrak{g}_\bullet^s$ are ($1$-Lipschitz) continuous when the domain is endowed with the $\mathcal{N}^s$-topology, so
$$\mathcal{N}^s\mbox{-topology } \supseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_2^s).$$
By Observation 4, if $0\in V$, so that the function that is constantly $0$ is in $C^r(U,V)$, then
$$\mathcal{T}(C^r(U,V),\mathfrak{F}_2^s)\subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s).$$
I believe the assumption that $V$ contains the origin is mild. Indeed, a general $V$ need not contain $0$, but since $V$ is supposed to be a chart of a manifold we could arrange it to contain $0$.
This final inclusion could be strict, provided that $V$ contains a nonzero $y$ and its negative $-y$. The point is that antipodes are not distinguishable by open sets coming from $\mathfrak{g}_\bullet^s$; see the discussion here.
Finally by Observation 5 we have that
$$\mathcal{N}^s\mbox{-topology }= \mathcal{T}(C^r(U,V),\mathfrak{F}_5^s),$$
which I think is the most interesting claim. In words, it says that the $\mathcal{N}^s$-topology on $C^r(U,V)$ is the smallest topology that makes the first $s$ derivatives continuous operators. An alternative reformulation by way of multiindex formalism is in Kelley & Namioka's Linear Topological Spaces, p. 82:
Putting all the inclusions and equalities together we have:
\begin{align*}
\mathcal{T}(C^r(U,V),\mathfrak{F}_1^s)
&\subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_2^s)\\
&\subseteq \mathcal{N}^s\mbox{-topology }
= \mathcal{T}(C^r(U,V),\mathfrak{F}_3^s)
= \mathcal{T}(C^r(U,V),\mathfrak{F}_5^s) \\
&\subseteq \mathcal{T}(C^r(U,V),\mathfrak{F}_4^s).
\end{align*}
(Of course on top of this we also have a grading w/r/t the order $s$ of derivatives; the higher the order of derivatives the smaller the open sets become. This would give refinements.)
