For the intersection of two plane curves (for simplicity, assumed to be defined over $\mathbb{C}$) at a point $P$, the multiplicity is computed as follows. Let $f$ be a local equation for one curve and $g$ a local equation for the other curve near the point $P$. Then multiplicity of $P$ in the intersection of the two curves is the dimension of the $\mathbb{C}$-vector space $\mathcal{O}_P / (f,g)$. Here $\mathcal{O}_P$ is the local ring of the plane at $P$ and $(f,g)$ is the ideal generated by $f$ and $g$.
In your example, consider the point $p = [0:1:0]$. We can work locally in the polynomial ring $\mathbb{C}[x,z]$, where $x = X/Y$ and $z = Z/Y$. (This is not quite the local ring $\mathcal{O}_P$ referred to above, but we can use it because it is the ring of functions in a neighborhood of $p$ that does not contain $r$.) The local equations for your curves are $f = x$ and $g = z^2 - x^4$. Thus, we must compute the dimension of $\mathbb{C}[x,z] / (x, z^2 - x^4)$, which has basis $\{1, z\}$. Thus the multiplicity of $p$ is $2$.