I think some of the commenters have forgotten the time when they found vector space linear algebra understandable, but triangulated categories confusing.
For someone in such a state, a useful tool to help understand statements about triangulated categories is passage to the Grothendieck group. Recall that this is done by taking the abelian group generated by the objects in the category $\mathcal{C}$, divided by the relation in which the sum of three elements of a triangle is zero. This gives an abelian group, and I'll tensor it with a field to get a vector space $[\mathcal{C}]$. Supposing in addition that Hom in the category gives finite dimensional vector spaces, $[\mathcal{C}]$ gets a bilinear form by $B(X, Y) = \dim Hom(X, Y)$.
For instance, in this case the thus "decategorified" statement is as follows:
Let $C$ and $D$ be vector spaces equipped with bilinear forms, and $F: C \to D$ a linear map. Say $X_i$ is a spanning set of vectors for $C$ and $F(X_i)$ spans $D$. Suppose $F$ is "an isometry on the spanning set", i.e., $B(X_i, X_j) = B(F(X_i), F(X_j))$. Then $F$ is an isometry.
This is a statement more accessible to intuition. And, as a proof of the original statement necessarily "decategorifies" to a proof of the decategorified one, often one can proceed in reverse and first prove the decategorified statement and then try to lift.
$\newcommand\Ocal{\mathcal{O}}\newcommand\Hom{\mathop{\mathrm{Hom}}}\newcommand\Ext{\mathop{\mathrm{Ext}}\nolimits}$I think the following is a rather clumsy way of proving it. It is an induction on two variables, but probably there are better ways to set up the induction. The idea is to prove it for
- $i$ between 0 and $n$, $j=0$
- $i=0$, $j$ between 0 and $n$
- show that the cases $(i,j)$ and $(i-1,j-1)$ are isomorphic
I will denote the sequence that Beilinson suggests to do the induction with by $(\#_i)$.
First part The case $i=0$ is clear. To see that $\Hom(\Omega^i(i),\Ocal)=\bigwedge^iV^*$ for $i\geq 1$ we apply $\Hom(-,\Ocal)$ to $(\#_i)$. One gets a four-term exact sequence
$$0\to\Hom(\Omega^{i-1}(i),\Ocal)\to\Hom(\bigwedge\nolimits^iV\otimes\Ocal,\Ocal)\to\Hom(\Omega^i(i),\Ocal)\to\Ext^1(\Omega^{i-1}(i),\Ocal)\to 0$$
and isomorphisms $\Ext^l(\Omega^i(i),\Ocal)\cong\Ext^{l+1}(\Omega^{i-1}(i),\Ocal)$ for $l\geq 1$. The second term in the sequence is isomorphic to $\bigwedge^i V^*$, which is what we are after. So we would like to have that $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$.
Lemma $\Ext^l(\Omega^{i-1}(i),\Ocal)=0$ for all $l$, and $i=1,\dotsc,n$.
Proof The case $i=1$ is clear. So assume that $i\geq 2$, and now we apply $\Hom(-,\Ocal(-1))$ to $(\#_{i-1})$. In the long exact sequence we get we see that $\Ext^l(\bigwedge\nolimits^{i-1}V^*\otimes\Ocal,\Ocal(-1))=0$ for all $l$. But then we apply the induction hypothesis, and conclude that everything is zero. $\square$.
So we get the desired isomorphism in this case.
Second part The case $j=0$ is clear. The case $j=n$ is also clear, because $\Omega^n(n)=\Ocal(-1)$. Then we rewrite the short exact sequence $(\#_{j+1})$ as
$$0\to\Omega^{j+1}(j)\to\bigwedge\nolimits^{j+1}\otimes\Ocal(-1)\to\Omega^j(j)\to 0$$
and then applying $\Hom(\Ocal,-)$ to this modified sequence gives a long exact sequence where $\Ext^l(\Ocal,\bigwedge\nolimits^{j+1}V\otimes\Ocal(-1))=0$, and similar to the previous lemma we can sandwich the desired term between zeroes.
Third part By now the ideas should be clear. Assume that $i\geq 1$ and $j\geq 1$. Apply $\Hom(-,\Omega^j(j))$ to $(\#_i)$. By the second part we see that the second (and fifth, etc.) term vanishes. We get another $\Ext^l=\Ext^{l+1}$ isomorphism.
Now apply $\Hom(\Omega^{i-1}(i),-)$ to $(\#_j)$. We get yet another long exact sequence, where by induction the third term $\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j))$ is isomorphic to $\bigwedge\nolimits^{i-j}V^*$ (and higher $\Ext^l$ vanish).
Now we do yet another induction on as in the lemma, to see that the second (and fifth, etc.) term vanishes. Hence tracing back all the isomorphisms we see that
\begin{align}
\bigwedge\nolimits^{i-j}V^*
&\cong\Hom(\Omega^{i-1}(i-1),\Omega^{j-1}(j-1)) \\
&\cong\Hom(\Omega^{i-1}(i),\Omega^{j-1}(j)) \\
&\cong\Ext^1(\Omega^{i-1}(i),\Omega^j(j)) \\
&\cong\Hom(\Omega^i(i),\Omega^j(j))
\end{align}
as desired. Hurray!
The geometric intuition should correspond to its interpretation as a dual exceptional collection, but I don't see a way to make this into an actual explanation. The above is just a proof by tedious and boring computation. I hope I didn't make mistakes.
Best Answer
Maps $\mathrm{B} G \to \mathcal{X}$ correspond to an object of $\mathcal{X}(k)$ along with a $G$-action. Indeed, the map $* \to \mathrm{B} G \to \mathcal{X}$ selects an object $x$ and for each test scheme $T$ we get a natural map,
$$ \{ T \text{-torsors} \} \to \mathcal{X}(T) $$
so that the trivial torsor maps to $x$. This is determined by the induced $G$-action on $x$ since, after a cover, each torsor becomes trivial and the gluing maps pass to gluing data for an object of $\mathcal{X}$.
Maps $[E/G] \to \mathcal{X}$ correspond to "$G$-equivariant objects over $E$" meaning the data of $x \in \mathcal{X}(E)$ with a $G$-action in $\mathcal{X}$ which is compatible with the $G$-action on $E$ along the functor $\mathcal{X} \to \mathrm{Sch}_k$.
First, $\mathbb{G}_m$-equivariant coherent sheaves are exactly coherent sheaves with a $\mathbb{Z}$-grading. In some sense, this is an incarnation of the fact that the actions of tori split up into a weight space decomposition. In the affine case, this is very clear, we need a comultiplication map,
$$ M \to M \otimes k[t, t^{-1}] $$
which gives a decomposition of $M$ into the sum over $M_n = \{ m \mid m \mapsto m \otimes t^n\}$. The (co)associativity of the action shows that this is a $\mathbb{Z}$-grading.
Now we need to consider $\mathbb{G}_m$-equivariant sheaves over $\mathbb{A}^1$. This means a coherent sheaf $\mathcal{F}$ on $X \times \mathbb{A}^1$ flat over $\mathbb{A}^1$ with a $\mathbb{G}_m$-action over $\mathbb{A}^1$. We get an actual sheaf $\mathcal{F}|_1$ taking the fiber over $1 \in \mathbb{A}^1$ which is what "$f(1)$" should mean. Consider this situation affine-locally. We have a $A[t]$-module $M$ which is $\mathbb{Z}$-graded compatibly with the $\mathbb{Z}$-grading on $A[t]$ (although this grading on $A[t]$ is trivial in negative degrees this need not be true of $M$ e.g. $M = A[t, t^{-1}]$. We think of $f(1) = M/(t-1)$ as the underlying object which messes up the grading but preserves the decreasing filtration
$$ M_{\ge n} = \bigoplus\limits_{k \ge n} M_n $$
because the action of $(t-1)$ preserves this filtration but not preserve the ascending filtration. Therefore, we get a $\mathbb{Z}$-filtered $A$-module $M/(t-1)$.
This is analogous to the "dynamical description of Parabolic, Levi, and Cartan subgroups" where we are asking for the limit of a $\mathbb{G}_m$-parametrized sheaf "as $t \to 0$". This is the sort of condition that gives a parabolic so it may not be so surprising that it recovers filtered objects.