The basic idea is to roll the triangle into a cone, though a little stretching is also involved. The top edge of the triangle (the intersection of $X$ with the line $y=1$) will become the circle at the top of the cone (the intersection of $Y$ with the plane $z=1$). More generally, each horizontal 'slice' through $X$, (i.e., each intersection of $X$ with a line $y=c$ for some $c$ with $0\le c\le 1$), will wrap around to become the corresponding circular horizontal 'slice' through $Y$ (i.e., the intersection of $Y$ with the plane $z=c$).
For $0\le c\le 1$ let $X_c=\{\langle x,y\rangle\in X:y=c\}$ and $Y_c=\{\langle x,y,z\rangle\in Y:z=c\}$; we want to map each $X_c$ to the corresponding $Y_c$. In doing this we'll obviously have to identity the endpoints of $X_c$, wrapping the line segment into a circle of radius $c$. The $x$-coordinates of points in $X_c$ range from $-c$ to $c$. One natural way to wrap the interval $[-c,c]$ into a circle of radius $c$ is to multiply it by $\frac{\pi}c$ (for $c>0$, of course!) and think of it as the $\theta$-coordinate in polar coordinates; the corresponding $x$- and $y$-coordinates are then $c\cos\frac{\pi}c$ and $c\sin\frac{\pi}c$, respectively. The resulting map is then
$f:X\to Y:\langle x,y\rangle\mapsto\begin{cases} \langle 0,0,0\rangle,&\text{if }y=0\\\\ \left\langle y\cos\frac{\pi}y,y\sin\frac{\pi}y,y\right\rangle,&\text{otherwise}\;. \end{cases}$
I'll leave it to you to check that this $f$ really is a quotient map.