The pair of pants is a cobordism realizing addition of circles

($P$ is a compact smooth surface with boundary the three circles.) Introduce the smooth functions ${\rho }_0(x)=4-\mathrm{\parallel }x{\mathrm{\parallel }}^2$, ${\rho }_{-}(x)=\mathrm{\parallel }x-(1,0){\mathrm{\parallel }}^2-\frac{1}{4}$ and ${\rho }_{+}(x)=\mathrm{\parallel }x+(1,0){\mathrm{\parallel }}^2-\frac{1}{4}$ on ${\mathbb{R}}^2$; then $P=\{x:{\rho }_0(x)\ge 0,{\rho }_{-}(x)\ge 0,{\rho }_{+}(x)\ge 0\}$. On $C_0=\{x:{\rho }_0(x)=0\}$ we have $\mathrm{\mid }x\mathrm{\mid }=2$, so $\mathrm{\parallel }x\mp (1,0)\mathrm{\parallel }\ge 2-1=1>\frac{1}{2}$ and the other two functions are strictly positive; on $C_{\pm }=\{x:{\rho }_{\pm }(x)=0\}$ we have $\mathrm{\parallel }x\pm (1,0)\mathrm{\parallel }=\frac{1}{2}$, so $\mathrm{\parallel }x\mathrm{\parallel }\le 1+\frac{1}{2}=\frac{3}{2}<2$ and, since the two inner centres are at distance $2$ and the radii sum to $1$, $\mathrm{\parallel }x\mp (1,0)\mathrm{\parallel }\ge 2-\frac{1}{2}=\frac{3}{2}>\frac{1}{2}$. Hence the three circles are pairwise disjoint and each boundary point of $P$ lies on exactly one of them, where exactly one defining function vanishes with nonzero gradient. Each such point therefore has a boundary chart obtained from that defining function by the inverse function theorem, so $P$ is a compact smooth surface with boundary $C_0\bigsqcup C_1\bigsqcup C_2$; compactness follows because $P$ is a closed subset of the compact disk $\{\mathrm{\mid }x\mathrm{\mid }\le 2\}$.

(The induced orientations of the three circles.) Give $P$ the orientation induced from the standard orientation of ${\mathbb{R}}^2$. At a point $p\in C_0$ the outward normal of $P$ is the radial unit vector $p\mathrm{/}2$, and $(p\mathrm{/}2,Jp\mathrm{/}2)$ is a positive basis of ${\mathbb{R}}^2$, where $J$ is the quarter-turn; by the outward-normal-first rule the positive tangent direction of $C_0$ is $Jp$, the counterclockwise direction. At a point $p\in C_{\pm }$, let $c$ be the centre of that circle; the removed disk lies outside $P$ in the direction $c-p$, so the outward normal of $P$ at $p$ is the unit vector $(c-p)\mathrm{/}\mathrm{\mid }c-p\mathrm{\mid }=2(c-p)$, and $(2(c-p),2J(c-p))$ is again a positive basis; hence the positive tangent direction is $J(c-p)=-J(p-c)$, which is the clockwise direction of the circle centred at $c$. So the induced orientation of the outer circle is counterclockwise and that of each inner circle is clockwise, i.e. the oriented boundary is $C_0-C_1-C_2=-(C_1\bigsqcup C_2)\bigsqcup C_0$.

($P$ is an oriented bordism from $C_1\bigsqcup C_2$ to $C_0$.) Take the incoming boundary part $(\mathrm{\partial }P{)}_0=C_1\bigsqcup C_2$ with the source orientations counterclockwise on both circles, and the outgoing part $(\mathrm{\partial }P{)}_1=C_0$; the induced orientations computed in step 2.1 are clockwise on $C_1,C_2$, which is the negative of the source orientation on the incoming part, and counterclockwise on $C_0$, which is the target orientation. The radial parametrisations ${\theta }_0(s,p)=c+(1+\frac{s}{4})(p-c)$ for $p$ in an inner circle with centre $c$, and ${\theta }_1(s,p)=(1+\frac{s}{4})p$ for $p\in C_0$, $s\in [0,1)$ respectively $s\in (-1,0]$, are smooth embeddings onto collar neighbourhoods of the corresponding boundary circles: their images have radii $\frac{1}{2}(1+\frac{s}{4})\in [\frac{1}{2},\frac{5}{8})$ and $2+\frac{s}{2}\in (\frac{3}{2},2]$ in the relevant radial directions and lie in $P$ by the estimates of step 1.1. Hence $P$ with these collars and this orientation is an oriented bordism from $C_1\bigsqcup C_2$ to $C_0$.

(The additive relation; all classes vanish.) By step 3.1 the cobordism class of $C_1\bigsqcup C_2$ equals that of $C_0$, that is, $[C_1]+[C_2]=[C_0]$ in ${\mathrm{\Omega }}_1^{SO}$ by [F4]. Each of the three circles is the boundary of a Euclidean disk (with the counterclockwise orientation induced by the standard orientation of the plane, after the outer circle is viewed as the boundary of the disk it encloses and each inner circle as the boundary of the removed disk), so by the disk example, which applies to a circle with either orientation, all three classes are zero in ${\mathrm{\Omega }}_1^{SO}$ and in ${\mathrm{\Omega }}_1^O$; the relation therefore reads $0+0=0$ and exhibits the disjoint-union addition of the group structure rather than an independent invariant.