Let $A = K[X,Y,Z]/(XZ-Y^2)$, $P = (\bar{X},\bar{Y}) \subseteq A$ an ideal and $M = A/P^2$ as an $A$-module. We know that $P^2 = (\bar{X}^2, \bar{Y}^2, \bar{X}\bar{Y}) = (\bar{X})\cap (\bar{X}, \bar{Y}, \bar{Z})^2$. Consider the submodules of $M$: $$N = \langle \bar{X} + P^{2}, \bar{Y} + P^{2}\rangle$$ $$L = \langle \bar{Y} + P^{2}\rangle$$ $$ F = \langle \bar{X} + P^{2}\rangle$$
Is it true that this isomorphism $$N/L \cong F$$ holds?
I cannot see why $L \cap F$ is trivial in $M.$
Note that $M \cong K[X,Y,Z]/(X^2,Y^2,XY,XZ)$. Your question is equivalent to asking why $(\bar{X}) \cap (\bar{Y})=0$ in this ring. Suppose $a \in (\bar{X}) \cap (\bar{Y})$. Then we may write $a=\bar{S}\bar{X}$ and $a=\bar{T}\bar{Y}$ for some $S,T \in K[X,Y,Z]$. In particular, this means $SX-TY \in (X^2,Y^2,XY,XZ)$. Thus $SX-TY=AX^2+BY^2+CXY+DXZ$ for some $A,B,C,D \in K[X,Y,Z]$. This forces $X$ to divide $BY^2+TY$ and so it must divide $BY+T$. Write $BY+T=UX$ for $U \in K[X,Y,Z]$. Then $SX=UXY+AX^2+CXY+DXZ$. In particular, $SX \in (X^2,Y^2,XY,XZ)$ and so $a=0$ in $M$.