思路如下：先找到两圆的交点，再分别画右边小圆的两段弧，左边一段弧画成虚线，右边一段弧正常即可。
注：1.使用let ... in ... 操作是为了找到两段弧的开始角度与结束角度。
2.使用atan反正切函数，注意此函数求出的角的范围是(-90,90)度。
3.使用库：intersections求交点坐标,calc实现坐标计算功能。

\documentclass[border=8pt,tikz]{standalone}
\usepackage{amssymb}
\usetikzlibrary{patterns,intersections,calc}
\begin{document} 

    \begin{tikzpicture}
        \path [draw, very thick, ->] node at (-.3,-.3) {$O$} (-0.5,0) -- (8,0) node[anchor=north] {$x$};
        \path [draw,very thick, ->](0,-0.5) -- (0,6.1)  node[anchor=east]  {$y$};
        \node  at (7.5,5.5) {$\mathbb{R}^{2}$};
        \draw[very thick,blue,dashed,-] (0.5,0.5) -- (6.5,0.5) -- (6.5,5) --(0.5,5)--(0.5,0.5);
        \draw[pattern=north west lines,name path=circle1,draw=red] (3,3) circle (1.5);
        \node [fill=white,rounded corners,below of=3] at (2.5,4) {$B$};
        \node at (2,1.3) {$ {\color{red}\partial B=B_{0}}$};
        \path[name path=circle2] (5,3) circle (1.2);
        %左边一段弧，dashed
        \draw[dashed,name intersections={of=circle1 and circle2,name=i}]
             let \p1=($(5,3)-(i-1)$), \p2=($(5,3)-(i-2)$), \n1={atan(\y1/\x1)}, \n2={atan(\y2/\x2)}
             in (i-1) arc({\n1+180}:{\n2+180}:1.2);
        %右边一段弧
        \draw let \p1=($(5,3)-(i-1)$), \p2=($(5,3)-(i-2)$), \n1={atan(\y1/\x1)}, \n2={atan(\y2/\x2)}
                 in     (i-1) arc({\n1+180}:{\n2-180}:1.2);
        \node at (6.2,1) {$\Lambda$};
        \node at (5.5,3) {$S$};
    \end{tikzpicture}  

谢谢你@u19850 了，也可以使用expl3宏包处理这一问题。

\documentclass{article}
\usepackage{fontspec}
\setmainfont{SimSun}
\usepackage{expl3}
\begin{document}

\ExplSyntaxOn
\int_step_inline:nnn {"3000} {"3FFF} {#1 \char #1 \quad }
\ExplSyntaxOff

\end{document}