Let's fix some notation for making things explicit. (The notation in your last 3 paragraphs is slightly confusing, so I'm making up my own). I take $A=V(z)$ and $B=V(y)$, hence $P=[1,0,0]$. Then $D_1=V(y^2z)$ and $D_2=V(yz^2)$, and hence
$$W=D_1 \cap D_2=V(y^2z,yz^2).$$
Since $D_i$ are divisors, we can calculate their normal bundles as $\mathcal{O}(D_i)|_{D_i} = \mathcal{O}(3)|_{D_i}$. In particular, after pulling back the normal bundle of $D_1\times D_2$ to the intersection, we get $N=\mathcal{O}(3)|_W\oplus \mathcal{O}(3)|_W$. The normal cone $C$ of $D_1\cap D_2$ to $\mathbb{P}^2$ sits inside $N$.
Let's restrict to the open dense subset of $W$ where $x\neq 0$ and calculate the cone and the irreducible components lying over this subset (one can easily see that we get nothing extra by considering the two points at infinity).
Then the base is $\mathrm{Spec}\ k[y,z]/(y^2z,yz^2)$ and the cone is, similar to what is written in the question,
$$C|_{D(x)}= \mathrm{Spec}\ (k[y,z]/(y^2z,yz^2) \oplus (y^2z,yz^2)/(y^2z,yz^2)^2 \oplus \dots).$$
Trivialize $N|_{D(x)}= \mathrm{Spec}\ k[y,z,s,t]/(y^2z,yz^2)$ with the obvious embedding given by $s\mapsto y^2z$ and $t\mapsto yz^2$. This is clearly a surjective map of rings, and the kernel is generated by $sz-yt$. (Remark: hence the description in the question is correct, after setting up everything such that it makes sense). We want to calculate the irreducible components of
$$\mathrm{Spec}\ k[y,z,s,t]/(y^2z,yz^2,sz-yt).$$
On the locus where $z\neq 0$, we have $y=s=0$, which gives us the $z$-$t$-plane minus the line $z=0$, and on the locus where $y\neq 0$, we get the $y$-$s$-plane, minus the line $y=0$. Taking the closure of these two, we get two two-dimensional irreducible components $C_1$ and $C_2$. Moreover, the fiber over $y=z=0$ is just $\mathrm{Spec}\ k[s,t]$, which is two-dimensional and irreducible and contained in the none of the other components, hence it is another irreducible component $C_3$. To conclude, the irreducible components of $C$ over $D(x)\subset W$ are given by the $z$-$t$-plane, the $y$-$s$-plane, and the $s$-$t$-plane.
This already gives us the distinguished varieties $A$, $B$ and $P$. To get the multiplicities, we need to calculate the length of the local rings.

The local rings at the first two components are just the quotient fields of the respective components (calculate on the complement of $y=z=0$). After localizing the coordinate ring at the prime ideal $(y,z)$, we get
$$k[y,z,s,t]/(y^2z,yz^2,sz-yt)_{(y,z)}=k(s,t)[y,z]/(y^2z,yz^2,z-\frac{t}{s}\cdot y),$$
which is the same as
$$k(s,t)[y]/(y^3).$$
This ring clearly has dimension $3$ as a $k(s,t)$-vector space, so by Example A.1.1 from Fulton, the last irreducible component has multiplicity $3$. Hence
$$[C]= [C_1]+[C_2]+3[C_3].$$

Then you apply Gysin for the vector bundle $N$ restricted to $A$, $B$, and $P$ to get the result. Note that $C_3$ is equal to the whole fiber of $N$ over $P$, so this simply gets mapped to the point $P$, since the vector bundle is trivial over the point $P$. On the other hand, the cycles $[C_1]$ and $[C_2]$ get mapped to zero-cycles of degree $3$ due to the fact that $N$ is a sum of two copies of $\mathcal{O}(3)$.