Suppose $R$ is a ring such that for every $R$-module $M$ there is a projective resolution $P_\bullet \to M$ of finite length (actually, this condition can be loosened quite a bit as I will note at the end). Then we can define a ring structure on $G(R)$ the Grothendieck group of finitely generated $R$-modules by defining the following product
$$
[A] \cdot [B] = \sum_i (-1)^i \text{Tor}_i(A,B).
$$
Note that if $P$ is a projective $R$-module, then $[A]\cdot[P] = [A \otimes P]$ since the $\text{Tor}_i(A,P)$ vanish for $i >0$.
The way to show this is a little painful. We need to show that
- This product is independent of choice of resolution,
- this product is well-defined,
- this product is symmetric,
- this product satisfies the ring axioms.
It is in my opinion easier to define this product as a map on $G(R)$ first, though tastes may vary. Fix a finitely generated $R$-module $M$ and let
$$
- \cdot M : G(R) \to G(R),\; [A] \mapsto [A] \cdot M := \sum_i (-1)^i [\text{Tor}_i(A,M)].
$$
Then one can show that this map is independent of the choice of resolution by showing $\text{Tor}_i(-,M)$ is. The well-definedness of the map follows since for any short exact sequende $0 \to A \to B \to C \to 0$ we have a long exact sequence of Tor, which shows $[A] \cdot M - [B] \cdot M + [C] \cdot M = 0$. The symmetry follows from the symmetry of Tor, so $[A] \cdot B = [B] \cdot A$.
From this point it is therefore acceptable to write $[A] \cdot [B]$, and we can prove the ring axioms one by one. For this it is good to note that for an $R$-module $M$ with projective resolution $P_\bullet \to M$ we have
$$
[M] \cdot [N] = \sum_i (-1)^i [\text{Tor}_i(M,N)] = \sum_i(-1)^i H_i(P_\bullet \otimes N) = \sum_i (-1)^i [P_i \otimes N].
$$
This computation involves the nontrivial lemma that in the Grothendieck group we have $\sum_i(-1)^i [C_i] = \sum_i (-1)^i [H_i(C_\bullet)]$ for any finite complex $C_\bullet$ in the category.
Then the ring axioms can easily be checked by using properties of the tensor product.
The requirement that every object in the category have a finite projective resolution is actually too strong. If every object has a finite resolution that is $F$-acyclic for a functor $F$ on the category, then it induces a derived morphism on Grothendieck groups. Actually, the resolution need not even be in the category itself. For example for categories of coherent sheaves we almost never have projective resolutions, but in some situations (details omitted on purpose) we may have locally free resolutions, which are $\text{Tor}_i(-, *)$ acyclic, so that we can define the product.
In general, one asks for a result like $U,V$ must intersect if something is true about their dimensions when $U,V$ are both closed inside some irreducible space - for instance, without requiring that they're closed, if they're both contained in lower-dimensional closed subvarieties and these subvarieties have no common components, you can often just delete the intersection points from one of $U$ or $V$ without changing anything about the circumstances of the problem. (The case of $U,V$ both open is trivial: a space is irreducible iff any two open subsets intersect, and any regular noetherian local ring is a domain, hence has irreducible spectrum, and any open subset of an irreducible space is again irreducible, so the punctured spectrum is again irreducible.)
In the case that $U,V$ are both closed, you know that they must intersect inside the punctured spectrum - there's only one closed point and every closed subset contains it, so every pair of closed subsets must intersect. This lets us get at the problem for the punctured spectrum in much the same manner as we treat intersections in projective space via the affine cone. The same proof will apply here - I leave it to you to verify the details.
Best Answer
This is just an expansion of my comment above. Giving a vector bundle on $X$ gives you a finitely generated module $M$ over $R$ such that its restriction to $X$ is your vector bundle. But, since $R$ is regular, you have a finite free resolution of $M$ over $R$ and thus restricting to $X$, you have a finite free resolution of your vector bundle. Thus, in the $K$ group, your vector bundle is just the alternate sum of finitely many trivial bundles and the rest should be clear.