This is more of a long comment:
Question 1: Shouldn't it be
$$f^*:\operatorname{Bun}_{r,d}(T)\to\operatorname{Bun}_{r,d}(S)?$$
No. You can't push vector bundles forward along general morphisms.
The functor is contravariant, as you point out, so we actually want $\operatorname{Bun}_{r,d}(-) : \mathsf{Sch}_k^\text{op} \rightarrow \mathsf{Set}$ which agrees with the direction the pull-back functor goes.
Question 2: What does this requirement actually mean? I.e. that a scheme represents a functor as above?
Yes it means exactly that, e.g. there is a scheme $M/{\operatorname{spec} k}$ such that the Yoneda functor $h_M = \mathrm{Hom}_{\mathsf{Sch}_k}(-,M)$ is naturally isomorphic to the functor $\operatorname{Bun}_{r,d}(-)$. That means that the set of maps of varieties from $X$ to $M$ exactly 'classifies' bundles of rank $r$ and degree $d$ on $X$. This idea, to my knowledge, originally comes from topology.
Question 3: How exactly can I understand this equality? It is not quite clear what the objects a morphism in both sides are and why there is some isomorphism between them.
The map comes from the Yoneda embedding and is essentially a very abstract tautology. We have a bijection which I'll lazily write as an equality:
$$\operatorname{Bun}_{r,d}(M) = \mathrm{Hom}_{\mathsf{Sch}_k}(M,M).$$
But the $\mathrm{Hom}$ set has a distinguished element, the identity $\mathrm{id}_M$! That means that — up to the ambiguity which might be involved in choosing a natural isomorphism — we have a canonical bundle $E_{r,d}/M$ of rank $r$ and degree $d$. Furthermore if you go back and carefully look through a proof of the Yoneda embedding and apply it in this situation, you'll find out that the above actually describes the function
$$\{\text{$\varphi:S\rightarrow M $ a $k$-morphism}\}=\{\text{$E$ vect. bundle of rank $r$ and degree $d$}\} / {\sim}$$
exactly as the operation $ \phi \mapsto \phi^* E_{r,d}$.
Question 4: Would you be able to clear this point out and explain it?
I'm not sure if I can do this one! This is quite a subtle point, and my own point of view is poisoned$^1$ by the higher categorical language.
The point is that the Yoneda functor obtained from a scheme is a contravariant functor to the category of sets, i.e. a pre-sheaf. When we want to classify objects with automorphisms, the naive way to do it will often fail to be a sheaf when one appropriately generalizes the notion of a sheaf to include such things. Instead what one can do is consider this as a sheaf of 'groupoids', that is a category whose morphisms are all isomorphisms.
A set naturally gives a groupoid by considering each element as an object with no morphisms besides the identity morphisms.
Classically, a sheaf is a presheaf $\mathcal{F}$ which satisfies for every open cover $\{U_i \}$ of an open set $U$ we have that the sequence
$$\mathcal{F}(U) \rightarrow \prod_{i} \mathcal{F}(U_i) \rightrightarrows \prod_{j,k} \mathcal{F}(U_j \cap U_k)$$
is an equalizer diagram (the analogue of a kernel for $\mathsf{Set}$).
The analogous notion for groupoids is not so simple. It involves not only the pairwise intersections $U_j \cap U_k$, but the triple-wise intersections $U_j \cap U_k \cap U_l$ and a cocycle condition on them.
Now for the poison: groupoids are equivalently$^2$ described as topological spaces who have a potentially non-trivial $\pi_0$ and $\pi_1$ but whose higher homotopy groups are trivial, i.e. $\pi_i=0$ for $i>1$.
A classical sheaf is then$^3$ a presheaf $\mathcal{G}$ of spaces which have possibly nontrivial $\pi_0$ but no higher homotopy groups, and we have that
$$\pi_0 \mathcal{G}(U) = \lim\Big\{ \prod_i \pi_0 \mathcal{G}(U_i) \rightrightarrows \prod_{j,k} \pi_0 \mathcal{G}(U_j \cap U_k) \Big\}.$$
To talk about sheaves of groupoids from a topological point of view we want that same above condition, but we want a condition relating the fundamental groups too! It is a general fact in topology that homotopy groups beyond $\pi_0$ do not behave well under limits in the category $\mathsf{Top}$, even fibered products can show this behavior.
$$\pi_1 \mathcal{G}(U) = \lim\Big\{ \prod_i \pi_1 \mathcal{G}(U_i) \rightrightarrows \prod_{j,k} \pi_1 \mathcal{G}(U_j \cap U_k) \overset{\rightarrow}{\underset{\rightarrow}{\rightarrow}} \prod_{s,r,t} \pi_1 \mathcal{G}(U_s \cap U_r \cap U_t)\Big\}.$$
This triple intersection cocycle condition is what guarantees this property of the $\pi_1$'s.
$^1$: I'm being facetious
$^2$: This is more subtle than I'm letting on
$^3$: See $^2$.
