$\newcommand\Ocal{\mathcal{O}}\newcommand\Hom{\mathop{\mathrm{Hom}}}\newcommand\Ext{\mathop{\mathrm{Ext}}\nolimits}$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.
I'm expanding on my comment, not really giving a complete answer.
Say that $L$ is a polarization on your smooth projective $X$. Then HRR applied to $F\otimes L^n$ gives
$$
P_L(F)(n)=\chi(F\otimes L^{\otimes n})=\int_X ch(F\otimes L^{\otimes n})\cdot td(X)=\int_X ch(F)\cdot e^{c_1(L)}\cdot td(X)
$$
where $P_L(F)$ is the Hilbert polynomial of $F$ with respect to $L$. So the Chern character recovers the Hilbert polynomial.
The Hilbert polynomial contains some information about your sheaf. For example its degree is the dimension of the support of $F$, and if $F$ is torsion-free (so that the degree of $P_L(F)$ is $dim(X)$), then, up to a constant term that depends only on $X$, the leading coefficient of $P_L(F)$ is the (generic) rank of $F$, and the next term gives you the degree of $F$.
If you didn't know already, you can find this stuff (and more) in the book "The geometry of moduli spaces of sheaves" by Huybrechts and Lehn.
I don't know about characterizing structure sheaves of subschemes on a K3 surface, but you have good chances of finding something in that same book (a large part of it is about sheaves on surfaces), or maybe someone else will comment here.
Best Answer
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.