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.