Drawing a cross using only one quadrant

Code from my friend.

% Bui Quy
\documentclass[tikz,border=5mm]{standalone}
\begin{document}
\begin{tikzpicture}[line join=round,line cap=round]
\def\edge{
(.5,.5)
to[out=90,in=-150] ++(1,2)
to[out=120,in=0] ++(-1,.5)
to[out=90,in=-30] ++(-.5,1)
to[out=-150,in=90] ++(-.5,-1)
to[out=180,in=60] ++(-1,-.5)
to[out=-30,in=90] ++(1,-2)
}

% Schrodinger's cat suggested this better code
\draw[fill=lime] (.5,.5) 
foreach \i in {0,90,180,270} {[rotate=\i]--\edge}
--cycle;
\end{tikzpicture}
\end{document}

here is my proposition, I modified your drawing and deleted the nodes and replaced by a drawing, it allows to use the mark changes and transformations

\documentclass{article}

\usepackage{verbatim}
\usepackage{tikz}
\begin{document}



\newcommand{\crossq}{

\def\coef{1}
\path[blue,thick,fill=red!30] (0,0) coordinate(O)
-- (180:0.6) coordinate(oual)
to [out=90,in=-10] coordinate[pos=0.23] (a1)++(-1,2.1)coordinate(a2)
to[out=50,in=180] ++(1,0.3)
to[out=90,in=-150] ++ (0.6,1)
-- cycle;

\draw[blue,thick] (a1)
to [out=90,in=-10] (a2)
to[out=50,in=180] ++(1,0.3)
to[out=90,in=-150] ++ (0.6,1);

\path[purple,thick,fill=red!30] (0,0) coordinate(O)
-- (0:0.6) coordinate(oual)
to [out=90,in=190] coordinate[pos=0.23] (b1)++(1,2.1)coordinate(b2)
to[out=130,in=0] ++(-1,0.3)
to[out=90,in=-30] ++ (-0.6,1)
-- cycle
;

\draw[purple,thick,fill=red!30] (b1)
to [out=90,in=190] (b2)
to[out=130,in=0] ++(-1,0.3)
to[out=90,in=-30] ++ (-0.6,1)
;

}



\begin{tikzpicture}
\begin{scope}
\crossq



\end{scope}
\begin{scope}[rotate=90]
\crossq
\end{scope}
\begin{scope}[rotate=-90]
\crossq
\end{scope}
\begin{scope}[rotate=180]
\crossq
\end{scope}

\end{tikzpicture}

\end{document}