There are a number of comments to make about Serre's intersection formula and its relation to derived algebraic geometry.
First, we should be a little more cautious about attribution. The idea of using "derived rings" to give an intrinsic version of the Serre intersection formula is not recent. The idea goes back at least to thoughts of Deligne, Kontsevich, Drinfeld, and Beilinson in the 1980s (and possibly earlier). These ideas have been made precise in a number of ways, in particular in work of Kapranov & Ciocan-Fontaine, and Toën & Vezzosi. EDIT: As Ben-Zvi reminded me below, one should also mention Behrend and Behrend-Fantechi on DG schemes and virtual fundamental classes. Of course Lurie's work has been the most comprehensive and powerful in its treatment of the foundations of DAG, but it's important to understand that his work arose in the context of these fascinating ideas.
Now, just to provide a little context, let me try to recall how Serre's formula arises from DAG considerations. Let's start by using the notation above, but let's assume for simplicity that $X$, $Y$, and $Z$ are all local schemes. (Some of the technicalities of DAG arise in making sheaf theory work with some sort of "derived rings," so our discussion will be easier if we ignore that for now.) So we write $X=\mathrm{Spec}(A)$, $Y=\mathrm{Spec}(B)$, and $Z=\mathrm{Spec}(C)$ for local rings $A$, $B$, and $C$.
Now if our aim is to intersect $Y$ and $Z$ in $X$, we know how to do that algebro-geometrically. We form the fiber product $Y\times_XZ=\mathrm{Spec}(B\otimes_AC)$. The tensor product that appears here is really the thing we're going to alter. To do that, we're going to regard $B$ and $C$ as (discrete) simplicial (commutative) $A$-algebras, and we're going to form the derived tensor product. This produces a new simplicial commutative ring $B\otimes^{\mathbf{L}}_AC$ whose homotopy groups are exactly the groups $\mathrm{Tor}^A_i(B,C)$. The intersection multiplicity is simply the length of $B\otimes^{\mathbf{L}}_AC$ as a simplicial $A$-module.
As Ben Webster says, the real joy of DAG is in thinking of the geometry of our new derived ring $B\otimes^{\mathbf{L}}_AC$ as a single unit instead of thinking only of its disembodied homotopy groups. The question you're asking seems to be: does thinking geometrically about this gadget help us to prove Serre's multiplicity conjectures in a more conceptual manner?
The short answer is: I don't know. I do not think a new proof of any of these has been announced using DAG (and it's definitely not in any of Lurie's papers), and in any case I do not think DAG has the potential to make the conjectures "easy." But let me see if I can make a case for the following idea: revisiting Serre's original method of reduction to the diagonal in the context of DAG.
Recall that, if $k$ is a field, if $A$ is a $k$-algebra, and if $M$ and $N$ are $A$-modules, then $$M\otimes_AN=A\otimes_{A\otimes_kA}(M\otimes_kN).$$
Hence to understand $\mathrm{Tor}^A_{\ast}(M,N)$, it suffices to understand $\mathrm{Tor}^{A\otimes_kA}_{\ast}(A,-)$. This allowed Serre to reduce to the case of the diagonal in $\mathrm{Spec}(A\otimes_kA)$. The key point here is that everything is flat over $k$, so Serre could only use this to prove the multiplicity conjectures for $A$ essentially of finite type over a field. Observe that the same equality holds if we work in the derived setting: if $M$ and $N$ are simplicial $A$-modules, and $A$ is an $R$-algebra, then the derived tensor product of $M$ and $N$ over $A$ can be computed as
$$A\otimes^{\mathbf{L}}_{A\otimes^{\mathbf{L}}_RA}(M\otimes^{\mathbf{L}}_RN).$$
The gadget on the right (or, strictly speaking, its homotopy) has a name familiar to toplogists; it's the Hochschild homology $\mathrm{HH}^R(A,M\otimes^{\mathbf{L}}_RN)$.
The hope is that we've chosen $R$ cleverly enough that $B\otimes^{\mathbf{L}}_RC$ is "less complicated" than $B\otimes^{\mathbf{L}}_AC$. (More precisely, we want the $\mathrm{Tor}$-amplitude of $M$ and $N$ to decrease when we think of them as $R$-modules. There's a particular way of building $R$, but let me skip over this point.)
Has our situation improved? Perhaps only a little: we've turned our problem of looking at the derived intersection $Y\times^h_XZ$ into the study of the derived intersection of the diagonal inside $X\times^h_RX$ with some simpler derived subscheme $Y\times^h_RZ$ thereof. But now we can try to iterate this, working inductively.
I don't know whether this can be made to work, of course.
I think it is difficult to compute the multiplicity just by looking "qualitatively" at the equations.
For instance, let $C \subset \mathbb{P}^3$ be a smooth curve over the field of complex numbers.
if $H \subset \mathbb{P}^3$ be a generic hyperplane. Then $H \cap C$ has intersection multiplicity $1$ at all points where it meets $C$.
let $c \in C$ be a generic point and let $H \subset \mathbb{P}^3$ be a generic hyperplane containing $T_{C,c}$. Then $H \cap C$ has multiplicity $2$ at $c$ and $1$ at the other points where it meets $C$ (the last part of this statement is non-trivial, it uses the Reflexivity Theorem in projective duality).
Of course one can distinguish easily between the first two cases because either $H$ is zero or not in $ \mathcal{M}_c/\mathcal{M}_c^2$ (where $\mathcal{M}_c$ is the maximal ideal of the point $c$).
On the other hand things become more ineteresting if you take $c_0 \in C$ such that $C$ has a flex point at $c_0$. Then, a general hyperplane containing $T_{C,c_0}$ has intersection multiplicity at least $3$ with $C$ at $c_0$.
Again things can be seen locally using higher differential spaces, but it becomes more and more tricky to explain what is going on. And here we have been dealing only with smooth space curves and linear spaces in $\mathbb{P}^3$. One can easily imagine that these intersection multiplicities are "very "hard" to compute in full generality.
Best Answer
There is no formula which looks only at the generic point(s) of $V \cap W$; you need to understand the entire sheaf $\mathcal{T}or_j^{\mathcal{O}_X}(\mathcal{O}_V, \mathcal{O}_W)$. It might be worth explaining the $K$-theory perspective on this.
Let $K_0(X)$ be the Grothendieck group of coherent sheaves on $X$. There is a ring structure on $K_0(X)$, where $$[\mathcal{E}] [\mathcal{F}] = \sum (-1)^j [\mathcal{T}or_j^{\mathcal{O}_X}(\mathcal{E}, \mathcal{F}) ]$$ for any coherent sheaves $\mathcal{E}$ and $\mathcal{F}$. Here I am using $[\mathcal{A}]$ to mean "class of $\mathcal{A}$ in $K_0(X)$", and I am using that $X$ is smooth to guarantee that the sum is finite.
$K_0(X)$ has a descending filtration, $K_0(X) \supseteq K_0(X)_{1} \supseteq K_0(X)_{2} \supseteq \cdots \supseteq (0)$ where $K_0(X)_i$ is spanned by classes of sheaves with support in codimension $i$. This makes $K_0(X)$ into a filtered ring, meaning that $$K_0(X)_i K_0(X)_j \subseteq K_0(X)_{i+j} \quad (\ast)$$ Containment $(\ast)$ is NOT obvious, and we will return to this point.
Let $gr \ K_0(X)$ be the associated graded ring $\bigoplus_{i \geq 0} K_0(X)_j/K_0(X)_{j+1}$. Then there is a map of graded rings from $gr \ K_0(X)$ to the Chow ring $A^{\bullet}(X)$. This map sends $[\mathcal{O}_V]$ to $[V]$.
So, let $V$ and $W$ live in codimensions $i$ and $j$. We want to compute $[V] [W]$ in $A^{i+j}(X)$. From the above, we see that it would be enough to compute $$\sum (-1)^j [\mathcal{T}or_j^{\mathcal{O}_X}(\mathcal{O}_V, \mathcal{O}_W) ] \quad (\ast \ast)$$ as an element of $K_0(X)_{i+j}/K_0(X)_{i+j+1}$.
Every summand in $(\ast \ast)$ is supported on $V \cap W$. So, if $V \cap W$ lives in codimension $i+j$, then we can just compute the image of each summand separately in the quotient $K_0(X)_{i+j}/K_0(X)_{i+j+1}$. Working this out gives Serre's formula.
Suppose now that $V \cap W$ has codimension $k$, which is less than $i+j$. Then the individual Tor terms live in $K_0(X)_k$ and plugging into Serre's formula gives the image of $(\ast \ast)$ in $K_0(X)_k/K_0(X)_{k+1}$. But, by containment $(\ast)$, the sum $(\ast \ast)$ actually lives farther down the filtration, in $K_0(X)_{i+j}$. This is why simply plugging into the formula you quote gives $0$.
An example might be useful. Take $X = \mathbb{P}^2$. Then $K_0(X)$ is isomorphic as an additive group to $\mathbb{Z}^3$, and we'll take as a basis the structure sheaf of $X$, the structure sheaf of a line, and the structure sheaf of a point. The filtration is given by $$(\ast, \ast, \ast) \supseteq (0, \ast, \ast) \supseteq (0,0,\ast) \supseteq (0,0,0)$$
Consider intersecting a line $V$ with itself. $\mathcal{T}or_0$ is the tensor product $\mathcal{O}_V \otimes \mathcal{O}_V$, whose class is $(0,1,0)$. $\mathcal{T}or_1$, is the restriction, to $V$, of the ideal sheaf of $V$. This is $\mathcal{O}_V(-1)$ and, as you can work out, it is $(0,1,-1)$ in the basis I chose. The other Tor terms are all zero.
So the individual Tor terms are $(0,1,0)$ and $(0,1,-1)$, which each live in $K_0(X)_1$ Those leading $1$ terms correspond to the lengths of the Tor modules at the generic point of $V$. In order to compute the intersection multiplicity, you have to see farther down in the filtration, to the element $(0,1,0) - (0,1,-1)$ in $K_0(X)_2$. Indeed, $(0,1,0) - (0,1,-1) = (0,0,1)$, showing that a line in the projective plane intersects itself in the class of a point.