Thinking about my comment again, maybe my remark on necessity of using left multiplication with $\Gamma$ was too hasty.

First, some general comment on the coset change in the specified situation. Assume that there is a matrix $A\in KZ$ such that $$ \left(\begin{array}{cc} t^3&bt^2+t^{-1}+lt\\0&1 \end{array}\right)= \left(\begin{array}{cc} t^3&\frac{y-m}{x-l}\\0&1 \end{array}\right)\cdot A $$ Then, working out the matrix multiplication, it is clear that $A=e_{12}(f)$ with
$$ ft^3=bt^2+t^{-1}+lt-\frac{y-m}{x-l}, f\in \Theta_\infty. $$ Rephrasing, the right coset representatives can be changed if and only if $bt^2+t^{-1}+lt$ and $\frac{y-m}{x-l}$ agree up to $t^3\Theta_\infty$. The right cosets are the same if and only if Takahashi's Proposition 1 can be applied to prove it.

However (and contrary to the formulation of the question), there are situations where that is actually possible (I think): let's look at the simplest example, which is $l=0$. In this case, we can start the Laurent series expansion at infinity: $\frac{y-m}{x}=t^{-1}-\frac{m}{x}$. As $x^{-1}=at^2$ with $a\neq 0$ up to terms of order $\geq 3$, we choose $m=-ba^{-1}$ and find that $$ \frac{y-m}{x}=t^{-1}+bt^2+ft^3 $$ for some $f\in\Theta_\infty$. In this case, it is in fact possible to change the right coset representative as required in Takahashi's paper.

[Added later:] In the general case, we make the following Laurent series expansion (correcting some mistakes in the reformulated question): $$ \frac{y-m}{x-l}=\frac{m-y}{l}\cdot\frac{1}{1-\frac{x}{l}}= \frac{m-y}{l}\left(-\sum_{n=1}^\infty\frac{1}{\left(\frac{x}{l}\right)^n}\right) =\frac{y-m}{l}\left(\sum_{n=1}^\infty\frac{l^n}{x^n}\right)=\left(t^{-1}-\frac{m}{x}\right)\sum_{n=0}^\infty\frac{l^n}{x^n}.$$

Now we write $x^{-1}=a_2t^2+a_3t^3+T_{geq 4}$, and expand the above product: $$ \left(t^{-1}-\frac{m}{x}\right)\sum_{n=0}^\infty\frac{l^n}{x^n}=\left(t^{-1}-\frac{m}{x}\right)\left(1+l\left(a_2t^2+a_3t^3+T_{\geq 4}\right)+R_{\geq 4}\right)=\left(t^{-1}+la_2t+la_3t^2+S'_{\geq 3}\right)-m\left(a_2t^2+S''_{\geq 3}\right) $$ with all the $T_{\geq 4}, R_{\geq 4},S'_{\geq 3},S''_{\geq 3}$ power series of the respective valuation. We collect the terms of order $\leq 2$ to get the following equality modulo $t^3\Theta_\infty$: $$ \frac{y-m}{x-l}=t^{-1}+la_2t+\left(la_3-ma_2\right)t^2. $$ I guess it is not a problem to choose $x$ such that $a_2=1$, and then the above allows to set $m=la_3-b$. This should show that also in the case $l\neq 0$, it is always possible to change the coset representative as required in Takahashi's paper.

Maybe the argument Takahashi had in mind was to see the case $l=0$ and then apply Proposition 6 of the paper to reduce the general case to the special case $l=0$.

Thinking about my comment again, maybe my remark on necessity of using left multiplication with $\Gamma$ was too hasty.

First, some general comment on the coset change in the specified situation. Assume that there is a matrix $A\in KZ$ such that $$ \left(\begin{array}{cc} t^3&bt^2+t^{-1}+lt\\0&1 \end{array}\right)= \left(\begin{array}{cc} t^3&\frac{y-m}{x-l}\\0&1 \end{array}\right)\cdot A $$ Then, working out the matrix multiplication, it is clear that $A=e_{12}(f)$ with
$$ ft^3=bt^2+t^{-1}+lt-\frac{y-m}{x-l}, f\in \Theta_\infty. $$ Rephrasing, the right coset representatives can be changed if and only if $bt^2+t^{-1}+lt$ and $\frac{y-m}{x-l}$ agree up to $t^3\Theta_\infty$. The right cosets are the same if and only if Takahashi's Proposition 1 can be applied to prove it.

There are situations where that is actually possible (I think): let's look at the simplest example, which is $l=0$. In this case, we can start the Laurent series expansion at infinity: $\frac{y-m}{x}=t^{-1}-\frac{m}{x}$. As $x^{-1}=at^2$ with $a\neq 0$ up to terms of order $\geq 3$, we choose $m=-ba^{-1}$ and find that $$ \frac{y-m}{x}=t^{-1}+bt^2+ft^3 $$ for some $f\in\Theta_\infty$. In this case, it is in fact possible to change the right coset representative as required in Takahashi's paper.

[Added later:] In the general case, we make the following Laurent series expansion (correcting some mistakes in the reformulated question): $$ \frac{y-m}{x-l}=\frac{m-y}{l}\cdot\frac{1}{1-\frac{x}{l}}= \frac{m-y}{l}\left(-\sum_{n=1}^\infty\frac{1}{\left(\frac{x}{l}\right)^n}\right) =$$ $$ =\frac{y-m}{l}\left(\sum_{n=1}^\infty\frac{l^n}{x^n}\right)=\left(t^{-1}-\frac{m}{x}\right)\sum_{n=0}^\infty\frac{l^n}{x^n}.$$

Now we write $x^{-1}=a_2t^2+a_3t^3+T_{\geq 4}$, and expand the above product: $$ \left(t^{-1}-\frac{m}{x}\right)\sum_{n=0}^\infty\frac{l^n}{x^n}=\left(t^{-1}-\frac{m}{x}\right)\left(1+l\left(a_2t^2+a_3t^3+T_{\geq 4}\right)+R_{\geq 4}\right)=\left(t^{-1}+la_2t+la_3t^2+S'_{\geq 3}\right)-m\left(a_2t^2+S''_{\geq 3}\right) $$ with all the $T_{\geq 4}, R_{\geq 4},S'_{\geq 3},S''_{\geq 3}$ power series of the respective valuation. We collect the terms of order $\leq 2$ to get the following equality modulo $t^3\Theta_\infty$: $$ \frac{y-m}{x-l}=t^{-1}+la_2t+\left(la_3-ma_2\right)t^2. $$ I guess it is not a problem to choose $x$ such that $a_2=1$, and then the above allows to set $m=la_3-b$. This should show that also in the case $l\neq 0$, it is always possible to change the coset representative as required in Takahashi's paper.

Maybe the argument Takahashi had in mind was to see the case $l=0$ and then apply Proposition 6 of the paper to reduce the general case to the special case $l=0$.

As an aside, maybe a bonus: if you are not necessarily interested in finding an explicitly embedded fundamental domain, you may get the computation of $\Gamma\backslash \mathcal{T}$ via Atiyah's classification of vector bundles on the elliptic curves. Originally, Atiyah's classification is for $k$ algebraically closed, but this has been generalized by Pumplün to arbitrary base fields as in Takahashi's setting. Then you are looking at the classification of vector bundles on the curve $\overline{E}=E\cup\{\infty\}$ which are trivial on $E$. The central point $o$ corresponds to the unique stable bundle with determinant at infinity. The points $v(l)$ form the $\mathbb{P}^1$ of semistable bundles of rank $2$ and degree $0$. All other points are direct sums of suitable line bundles. The stabilizer of the vertex in the tree corresponds to the automorphism group of the vector bundle (all this is in Serre's book on trees). You can work out the action of the automorphism groups on the links, compute the quotients of links modulo the automorphism group actions and glue these local models together - you get exactly Takahashi's description of $\Gamma\backslash\mathcal{T}$. I find this way of argument a bit simpler than the explicit calculations in Takahashi's paper (this way I worked out the next case, the quotient of the two-dimensional building modulo $GL_3(k[E])$).

Thinking about my comment again, maybe my remark on necessity of using left multiplication with $\Gamma$ was too hasty.

First, some general comment on the coset change in the specified situation. Assume that there is a matrix $A\in KZ$ such that $$ \left(\begin{array}{cc} t^3&bt^2+t^{-1}+lt\\0&1 \end{array}\right)= \left(\begin{array}{cc} t^3&\frac{y-m}{x-l}\\0&1 \end{array}\right)\cdot A $$ Then, working out the matrix multiplication, it is clear that $A=e_{12}(f)$ with
$$ ft^3=bt^2+t^{-1}+lt-\frac{y-m}{x-l}, f\in \Theta_\infty. $$ Rephrasing, the right coset representatives can be changed if and only if $bt^2+t^{-1}+lt$ and $\frac{y-m}{x-l}$ agree up to $t^3\Theta_\infty$. The right cosets are the same if and only if Takahashi's Proposition 1 can be applied to prove it.

However (and contrary to the formulation of the question), there are situations where that is actually possible (I think): let's look at the simplest example, which is $l=0$. In this case, we can start the Laurent series expansion at infinity: $\frac{y-m}{x}=t^{-1}-\frac{m}{x}$. As $x^{-1}=at^2$ with $a\neq 0$ up to terms of order $\geq 3$, we choose $m=-ba^{-1}$ and find that $$ \frac{y-m}{x}=t^{-1}+bt^2+ft^3 $$ for some $f\in\Theta_\infty$. In this case, it is in fact possible to change the right coset representative as required in Takahashi's paper.

I am not entirely certain about the general case, because I did not do the Laurent series expansion.

Thinking about my comment again, maybe my remark on necessity of using left multiplication with $\Gamma$ was too hasty.

First, some general comment on the coset change in the specified situation. Assume that there is a matrix $A\in KZ$ such that $$ \left(\begin{array}{cc} t^3&bt^2+t^{-1}+lt\\0&1 \end{array}\right)= \left(\begin{array}{cc} t^3&\frac{y-m}{x-l}\\0&1 \end{array}\right)\cdot A $$ Then, working out the matrix multiplication, it is clear that $A=e_{12}(f)$ with
$$ ft^3=bt^2+t^{-1}+lt-\frac{y-m}{x-l}, f\in \Theta_\infty. $$ Rephrasing, the right coset representatives can be changed if and only if $bt^2+t^{-1}+lt$ and $\frac{y-m}{x-l}$ agree up to $t^3\Theta_\infty$. The right cosets are the same if and only if Takahashi's Proposition 1 can be applied to prove it.

However (and contrary to the formulation of the question), there are situations where that is actually possible (I think): let's look at the simplest example, which is $l=0$. In this case, we can start the Laurent series expansion at infinity: $\frac{y-m}{x}=t^{-1}-\frac{m}{x}$. As $x^{-1}=at^2$ with $a\neq 0$ up to terms of order $\geq 3$, we choose $m=-ba^{-1}$ and find that $$ \frac{y-m}{x}=t^{-1}+bt^2+ft^3 $$ for some $f\in\Theta_\infty$. In this case, it is in fact possible to change the right coset representative as required in Takahashi's paper.

[Added later:] In the general case, we make the following Laurent series expansion (correcting some mistakes in the reformulated question): $$ \frac{y-m}{x-l}=\frac{m-y}{l}\cdot\frac{1}{1-\frac{x}{l}}= \frac{m-y}{l}\left(-\sum_{n=1}^\infty\frac{1}{\left(\frac{x}{l}\right)^n}\right) =\frac{y-m}{l}\left(\sum_{n=1}^\infty\frac{l^n}{x^n}\right)=\left(t^{-1}-\frac{m}{x}\right)\sum_{n=0}^\infty\frac{l^n}{x^n}.$$

Now we write $x^{-1}=a_2t^2+a_3t^3+T_{geq 4}$, and expand the above product: $$ \left(t^{-1}-\frac{m}{x}\right)\sum_{n=0}^\infty\frac{l^n}{x^n}=\left(t^{-1}-\frac{m}{x}\right)\left(1+l\left(a_2t^2+a_3t^3+T_{\geq 4}\right)+R_{\geq 4}\right)=\left(t^{-1}+la_2t+la_3t^2+S'_{\geq 3}\right)-m\left(a_2t^2+S''_{\geq 3}\right) $$ with all the $T_{\geq 4}, R_{\geq 4},S'_{\geq 3},S''_{\geq 3}$ power series of the respective valuation. We collect the terms of order $\leq 2$ to get the following equality modulo $t^3\Theta_\infty$: $$ \frac{y-m}{x-l}=t^{-1}+la_2t+\left(la_3-ma_2\right)t^2. $$ I guess it is not a problem to choose $x$ such that $a_2=1$, and then the above allows to set $m=la_3-b$. This should show that also in the case $l\neq 0$, it is always possible to change the coset representative as required in Takahashi's paper.

Maybe the argument Takahashi had in mind was to see the case $l=0$ and then apply Proposition 6 of the paper to reduce the general case to the special case $l=0$.