I used Van-Kampen theorem to show that $\pi_1(K^2) = \langle a,b : ab=b^{-1}a\rangle$ then I used Van-Kampen again to split $K^2$ to two Mebius strips and when taking the amalgamation with the intersection of the Mebius strips I got the free group $\langle a,b | a^2=b^2\rangle$.
I am trying to find the direct isomorphism - for $a$ I got:
$ab=b^{-1}a \Rightarrow a=b^{-1}ab^{-1} \Rightarrow a^2=a(b^{-1}ab^{-1}) = (ab^{-1})^2$ so I can define the isomorphism on $a \rightarrow(ab^{-1})$.
But I don't succeed to find an expression the $b^2$, I was trying to search in the polygonal representation of $K^2$ but could't make it. I would really appreciate an answer which gives a motivation/intuition for $b^2$ expression from $K^2$ characteristics.
Note that $ab=b^{-1}a$ implies $bab=a$ and so suggests $(ab)^2=abab=a^2$ as something to build the isomorphism on. Indeed, we check that $$\begin{align}\phi\colon \langle \,a,b\mid a^2=b^2\,\rangle&\to \langle \,a,b\mid ab=b^{-1}a\,\}\\ a&\mapsto a\\b&\mapsto ab\end{align}$$ and $$\begin{align}\psi\colon \langle \,a,b\mid ab=b^{-1}a\,\rangle&\to \langle \,a,b\mid a^2=b^2\,\}\\ a&\mapsto a\\b&\mapsto a^{-1}b\end{align}$$ are well-defined because $$\phi(a^2)=a^2=abab=\phi(b)^2=\phi(b^2)$$ and $$\psi(ab)=a a^{-1}b=b =b^{-1}b^2=b^{-1}a^2 =b^{-1}a\cdot a=\psi(b)^{-1}\psi(a)=\psi(b^{-1}a) $$ and are inverse of each other because $$\psi(\phi(a))=\psi(a)=a,\qquad \psi(\phi(b))=\psi(ab)=aa^{-1}b=b$$ and $$\phi(\psi(a)=\phi(a)=a\qquad \phi(\psi(b))=\phi(a^{-1}b)=a^{-1}ab=b. $$ We conclude that $\phi$ is an isomorphism (with inverse $\psi$).