Let $K_0=\{a,b\}$ and let $K_1=\{X,Y\}$ with $\partial_0(X)=\partial_1(Y)=a$ and $\partial_1(X)=\partial_0(Y)=b$. This is a $\Delta$-complex.
The space $\bar{K}$ is a disjoint union of two discrete points given by $a$ and $b$, and two disjoint intervals given by $X$ and $Y$. The equivalence relation just says that we join the beginning of the $X$ interval and the end of the $Y$ interval to the point $a$, and the end of the $X$ interval and the beginning of the $Y$ interval to the point $b$. Draw a picture of this and you will see that the geometric realisation $|K|=\bar{K}/{\sim}$ is homeomorphic to a circle.
If we instead want to represent the circle by a simplicial complex $K$, then we need at least three $1$-simplices. Let $K_0=\{a,b,c\}$ and let $K_1 = \{\{a,b\},\{b,c\},\{c,a\}\}$ with $$\partial_0(\{a,b\})=\partial_1(\{c,a\})=a\\ \partial_0(\{b,c\}) = \partial_1(\{a,b\})=b\\ \partial_0(\{c,a\}) = \partial_1(\{b,c\})=c$$
and this time we have three edges which are glued together in the geometric realisation according to the ordering of their boundary faces given by the image of the $\partial_i$.