Take the exercise from Arnold's book at page 10 where we are told to solve the singularity at $0$ of the curve $x^2=y^3$. The solution is given by the following graphs :
From the first graph, we consider the space $\mathbb{R}P^1\times \mathbb{R}$ with local coordinates $(x/y,y)$ and the curve has the following equation : $v^2=y$, hence we obtain the second graph.
But then how to obtain the third graph since we know that the image a contact of order $1$ by a blow up is a contact of order $0$. So we should have only two straight lines intersecting at $0$ ? why $3$ ?
I don't understand what's going on. Ordinarily we look at just the proper transform of the curve when we blow up, not the total transform. The equation in $(v,y)$-space becomes $(vy)^2=y^3$, so $y^2=0$ or $y=v^2$. The proper transform is just the parabola $y=v^2$, which is already smooth. If we are going to include $y^2=0$, then we have the parabola tangent to a double line. If we blow up again by setting $y=vz$, then we have $(vz)^2=0$, $v=0$, and $z=v$, so we have three lines (two double).
Can you edit your post to include more details from the text. Nor do I understand your "contact of order 1" remark. If we have a parabola and a line, it's then a double line, so we have contact order 2?