I am reading a paper on Euclidian cones (https://arxiv.org/abs/1103.0197), and there's this example (2.7):
Let $M:=\big(\frac{1}{\sqrt{3}}\mathbb{S}^2\big)\times\big(\frac{1}{\sqrt{3}}\mathbb{S}^2)$. Let then $u_1, u_2$ be an orthonormal basis of of $T_x\big(\frac{1}{\sqrt{3}}\mathbb{S}^2\big)$ and $v_1,v_2$ analogously for $T_y\big(\frac{1}{\sqrt{3}}\mathbb{S}^2\big)$. Then an orthonormal basis of $T_{(x,y)}M$ is given by $\{\bar{u_1},\bar{u_2},\bar{v_1},\bar{v_2}\}$ with $\bar{u_i}=(u_i,0)$ and $\bar{v_i}=(0,v_i)$. In this basis:$$Sec_{(x,y)}(\bar{u_1},\bar{u_2})=Sec_{(x,y)}(\bar{v_1},\bar{v_2})=3$$and $0$ in all the other cases so for each $\xi\in T_{(x,y)}M$ we have $Ric_{(x,y)}(\xi,\xi)=3$ (this also appears in the Petersen's book).
But now comes the thing that I do not understand: let $N:=(M\times[0,\infty)/M\times\{0\})\backslash\{O\}$ (you identify the points with second coordinate $0$ and remove the origin to get a "punctured cone") so that given $r>0$ an orthonormal basis of $T_{(x,y,r)}N$ is given by $\{\hat{u_1},\hat{u_2},\hat{v_1},\hat{v_2},\hat{w}\}$ with $\hat{u_i}=\frac{1}{r}(u_i,0,0)$, $\hat{v_i}=\frac{1}{r}(0,v_i,0)$ and $\hat{w}=(0,0,1)$. In this basis they claim that$$Sec_{(x,y,r)}(\hat{u_1},\hat{u_2})=\frac{2}{r^2},\quad Sec_{(x,y,r)}(\hat{u_1},\hat{v_1})=-\frac{1}{r^2}=Sec_{(x,y,r)}(\hat{u_1},\hat{v_2}),\quad Sec_{(x,y,r)}(\hat{u_1},\hat{w})=0.$$So now the Ricci is $0$ but what I do not understand is why the scalar curvatures changed from $0$ to $-\frac{1}{r^2}$ and the first one to $\frac{2}{r^2}$.