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.
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.
Best Answer
Consider a flat morphism $f:X\to Y$ of smooth connected varieties. For instance let $X=Y\times F$ with $X,Y,F$ all smooth. Further let $Z\subset X$ be a generically reduced subvariety such that $f|_Z:Z\to Y$ is a finite morphism, which is not flat. For any $y\in Y$ let $X_y\subset X$ denote the fiber of the original $f$.
As any two points on $Y$ are equivalent, the intersection product $Z\cdot_X X_y$. is independent of $y\in Y$.
Since $Y$ is smooth and $Z$ is generically reduced, most fibers of $f|_Z$ are smooth, so the intersection product $Z\cdot_XX_{y}$ for a general $y\in Y$ is given by the length of $\mathscr O_{X_{y}}$, in other words, there is no $\mathrm{Tor}$ contribution there.
Now take a point $y\in Y$ such that $f_Z$ is not flat in any neighbourhood of $y$ in $Y$. It follows that the length of $\mathscr O_{X_y}$ (which is also the Hilbert polynomial of the fiber) has to be different from the value that we get from the smooth fibers. Therefore there has to be $\mathrm{Tor}$ contribution there.
Perhaps this example shows why it is $\mathrm{Tor}$ that comes in: $\mathscr O_{Z_y}$ is not flat on $Z$, so it does not give the right length and thus has to be corrected. It is arguably intuitive that the alternating sum of the length of the $\mathrm{Tor}$'s will be constant as $y$ runs through the (closed) points of $Y$.
I don't know if you find this geometrically satisfying. I would paraphrase what Hailong said and say that anything involving $\mathrm{Tor}$ should not be geometrically obvious, on the other hand $\mathrm{Tor}$ measures the failure of exactness of the tensor product so the failure of flatness is an obvious condition to look at. Then again, all of this just reiterates the fact that flatness is a difficult notion to truly comprehend (and I don't claim I do).
The examples/comments given by JC and Hailong fit into this example:
Since $Y$ is smooth and $f|_Z$ is finite, it is flat if and only if $Z$ is Cohen-Macaulay, which gives us Hailong's comment, and
a concrete example is given by $X=\mathbb A^4$, $Y=\mathbb A^2$, $f$ the obvious projection, and $Z$ the union of two planes meeting in a single point both of which project onto $Y$ isomorphically, which gives us JC's example.