Total and strict transform of a line through the origin

In the first chart of [F4] write $s=y\mathrm{/}x$, so the chart ring is $k[x,s]$ with $y=xs$ and the exceptional curve is $E=V(x)$; the line $L$ has local equation $f=y$ at the origin, and $f(x,xs)=xs=x\cdot s$ with $x\nmid s$, so by [F5] the total transform is cut by $x\cdot s$ and the strict transform is cut by $s$, the divided equation of multiplicity $m=1$. The preimage of $L\setminus \{0\}$ is the locus $\{s=0,x\mathrm{\ne }0\}$ of this chart, whose closure is $L^{\mathrm{\prime }}=V(s)\cong \mathrm{Spec}k[x]$, and $E=V(x)$; hence ${\pi }^{\ast }L=(x)+(s)=E+L^{\mathrm{\prime }}$ as effective Cartier divisors, in agreement with [F3], and $\pi$ restricts on $L^{\mathrm{\prime }}=V(s)$ to $(x,s)\mapsto (x,xs)=(x,0)$, an isomorphism onto $L$.

In the second chart of [F4] write $t=x\mathrm{/}y$, so the chart ring is $k[y,t]$ with $x=yt$ and $E=V(y)$; the line $L$ is $V(y)$, whose pullback there is $E$ itself with no residual factor, so no point of the strict transform lies in this chart. Indeed $(y):y^{\mathrm{\infty }}=(1)$, so the defining saturated ideal is the unit ideal and the strict-transform chart is empty.

The two chart computations glue: steps 1.1 and 2.1 give $L^{\mathrm{\prime }}$ as the closed subscheme cut by $s$ in the first chart and by the unit ideal in the second, and these descriptions agree on the overlap, where $L^{\mathrm{\prime }}$ is empty; hence $L^{\mathrm{\prime }}$ is the closure of the preimage of $L\setminus \{0\}$ and is isomorphic to $L$ through $\pi$. In the first chart $L^{\mathrm{\prime }}=V(s)$ and $E=V(x)$ meet in the single reduced point $V(x,s)$, the point of $E$ with chart coordinate $s=0$, which is exactly the direction of $L$; the two curves are the coordinate axes there, so they are regular with distinct tangent lines and meet transversally with contact order one.

Collecting the results: ${\pi }^{\ast }L=L^{\mathrm{\prime }}+E$ with multiplicity one along $E$, the strict transform $L^{\mathrm{\prime }}$ is isomorphic to $L$ and meets $E$ transversally in the single point of $E$ corresponding to the direction of $L$, the strict transform has no points in the second chart, and the total transform $L^{\mathrm{\prime }}\cup E$ is the full preimage of $L$. This is exactly the assertion of the statement.