OK. Let's try this version which at least (I think) answers the special case.
Consider the map $\Psi_k$ sending a polynomial $h$ to $D^k(\Phi(h))$ (where $D$ represents differentiation with respect to $t$). We have that $M_{k-1}$ is precisely the kernel of $\Psi_k$. Notice that $\Psi_k$ is a group homomorphism (but not a ring homomorphism). In the case where $f_1$ and $f_2$ are linear functions, they satisfy the important property $D^2f_i=0$. Notice that $Df_1=1$ and $Df_2=\sqrt{2}$.
Now I claim that $$ D^k \Phi(h) = \sum_{i+j=k}\binom{k}{i\ j}\frac{\partial^k h}{\partial x_1^i\partial x_2^j}(f_1(t),f_2(t)). (Df_1)^i(Df_2)^j. $$ This can be rewritten as $$ \sum_{i+j=k}\binom{k}{i\ j}\Phi\left(\frac{\partial^k h}{\partial x_1^i\partial x_2^j}\right)\sqrt{2}^j. $$ Since $\Phi$ is a $\mathbb C$-linear group homomorphism, this is the same as $$ \Phi\left(\sum_{i\le k}\binom {k}{i} \sqrt{2}^{k-i}\frac{\partial^k h}{\partial x_1^i\partial x_2^{k-i}}\right) $$ Let $A_k(h)$ denote the element of $\mathbb Q[\sqrt 2][x_1,x_2]$ inside the parentheses. This may be succinctly written as $$ A_k(h)=\left(\frac{\partial}{\partial x_1}+\sqrt{2}\frac{\partial}{\partial x_2}\right)^kh $$
I claim that $f_1$ and $f_2$ are algebraically independent over $\mathbb Q[\sqrt 2]$. To see this consider the lexicographic ordering on monomials $x_1^ax_2^b$ where $x_1^ax_2^b\succ x_1^{a'}x_2^{b'}$ if $b>b'$ or $b=b'$ and $a>a'$. If $p\in \mathbb Q[\sqrt 2][x_1,x_2]$ is a relation on $f_1$ and $f_2$: $p(f_1,f_2)=0$, consider the lexicographically maximal term in $p$. This yields a term in $p(f_1,f_2)$ of the form $cr^at^b$ where $c\in \mathbb Q[\sqrt 2]$. This cannot be cancelled by any lower term in $p$.
We showed above that $D^k\Phi(h)=\Phi(A_k(h))$ so since the kernel of $\Phi$ is trivial we see that $M_k$ is equal to the kernel of $A_k$ lying inside $\mathbb Z[x_1,x_2]$. To compute the kernel of $A_k$, we define new monomials $u=x_1+x_2/\sqrt{2}$ and $v=x_1-x_2/\sqrt 2$. We calculate $\frac{\partial}{\partial u}=\frac12(\frac\partial{\partial x_1}+\sqrt 2\frac\partial{\partial x_2})$ and $\frac{\partial}{\partial v}=\frac12(\frac\partial{\partial x_1}-\sqrt 2\frac\partial{\partial x_2})$. The operator $A_k$ is precisely $2^k\partial^k/\partial u^k$ so we see that the kernel of $A_k$ (lying in $\mathbb Q[\sqrt 2][x_1,x_2]$) is generated by the elements of the form $u^iv^j$ for $0\le i < k$ and $j$ arbitrary and with coefficients in $\mathbb Q[\sqrt 2]$. We now need the elements of this group that lie in $\mathbb Z[x_1,x_2]$.
Let $P(u,v)=\sum_{i,j}a_{i,j}u^iv^j$ be a polynomial in $u$ and $v$ (with coefficients in $\mathbb Q[\sqrt 2]$. It lies in $\mathbb Q[x_1,x_2]$ if and only if it is fixed by the field automorphism $\theta$ of $\mathbb Q[\sqrt 2]$ switching $\sqrt 2$ and $-\sqrt 2$. Since $\theta(u)=v$ and $\theta(v)=u$, this means that we are requiring $a_{j,i}=\theta(a_{i,j})$. The elements of $M_{k-1}$ lying in $\mathbb Q[x_1,x_2]$ are spanned by $\sqrt{2}(u^iv^j-u^jv^i)$ for $i$ and $j$ satisfying $0\le i < j < k$ and $u^iv^j+u^jv^i$ for $i$ and $j$ satisfying $0\le i\le j < k$. It follows that $M_{k-1}$ is $k^2$-dimensional.