= Solution
For $M>a\ge0$ and $N>b\ge0$, define
$$
\boxed{K_{a,b}^{M,N}=\langle x^M,y^N,t(a,b)\rangle\le K.}
$$
The <normal form theorem for a free product> gives
$$
K_{a,b}^{M,N}\cong\langle X,Y,T\mid XY=YX\rangle\cong\mathbb Z^2*\mathbb Z,
$$
where $X,Y,T$ map to $x^M,y^N,t(a,b)$. To see injectivity, expand an alternating word in $\langle x^M,y^N\rangle$ and powers of $t(a,b)$. The conjugating $x,y$ factors commute with the lattice syllables, so each intervening nonidentity lattice syllable stays nonidentity in the expanded free-product normal form.
Its intersection with $T_K$ is the <free group> with basis
$$
K_{a,b}^{M,N}\cap T_K
=\langle t(Mu+a,Nv+b):u,v\in\mathbb Z\rangle.
$$
This follows by moving all $x^M,y^N$ factors to the right; the zero-exponent lattice remainder characterizes membership in the kernel.
The required <isomorphisms> are specified on the three generators by
$$
\boxed{\phi_i:x^m\mapsto x^{m^2},\quad y^m\mapsto y,\quad t(a_i,b_i)\mapsto t(c_i,0),}
$$
and
$$
\boxed{\varphi_j:x^m\mapsto x,\quad y^m\mapsto y^{m^2},\quad t(a_j,b_j)\mapsto t(0,c_j).}
$$
Each source and target has the same presentation $\langle X,Y,T\mid[X,Y]=1\rangle$, and each displayed assignment identifies its three abstract generators. Hence these are genuine <isomorphisms>, not merely maps of generating sets.
On the kernel basis they give exactly the machine transitions:
$$
\phi_i\bigl(t(mu+a_i,mv+b_i)\bigr)=t(m^2u+c_i,v),
$$
$$
\varphi_j\bigl(t(mu+a_j,mv+b_j)\bigr)=t(u,m^2v+c_j).
$$
For example, rewrite the first input as $x^{-mu}y^{-mv}t(a_i,b_i)x^{mu}y^{mv}$ and apply $\phi_i$ to each factor. These identities also hold for negative $u,v$, since they are identities in the <groups>.
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