$GL_2(\mathbb{C})$-invariant ring for $M_2(\mathbb{C})\times M_2(\mathbb{C})$ For $\mathbb{C}^*$-action on $\mathbb{C}^2$ by $t\circ(x,y)=(t^{-1}x,ty)$, the ring of invariant polynomials is $\mathbb{C}[xy]$. 
For $\mathbb{C}^*$-action on $\mathbb{C}^4$ by $t\circ(x,y,z,w)=(t^{-1}x,t^{-1}y,tz,tw)$, then the ring of invariants is $\mathbb{C}[xz,xw,yz,yw]$, which is a quadratic cone since there is a relation among the generators: $\mathbb{C}[X,Y,Z,W]/\langle XW-YZ\rangle$. 
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For $G = GL_2(\mathbb{C})$ acting on $M_2(\mathbb{C})$ by conjugation ($g\circ m=gmg^{-1}$), the ring of invariants is $\mathbb{C}[\det(m),\textrm{tr}(m)]$. 
For  $G$ acting on $M_2(\mathbb{C})\times M_2(\mathbb{C})$ by $g\circ (m,n)=(gmg^{-1},gng^{-1})$, why is it that the ring of invariant polynomials under the group action is only $\mathbb{C}[\textrm{tr}(m),\det(m),\textrm{tr}(n),\det(n)]$? More precisely, why doesn't there exist other $G$-invariants using a combination of $m$ and $n$? 
If we restrict $G$-action to $S= \{ (m,n): mn=nm\} \subset M_2\times M_2$, then is it true that $\mathbb{C}[S]^{G} = \mathbb{C}[\textrm{tr}(m),\det(m),\textrm{tr}(n),\det(n)]$?
 A: Your statement is wrong. Consider the function "Coefficient of $xy$ in $\det(x M + y N)$". Explicitly, this is $m_{11} n_{22} + m_{22} n_{11} - m_{12} n_{21} - m_{21} n_{12}$. From the first description, it is clearly invariant for the action $g \circ (m,n) \mapsto (g m g^{-1}, g n g^{-1})$. 
I claim that this is NOT in the ring generated by $Tr(M)$, $Tr(N)$, $Det(M)$ and $Det(N)$. Proof: The four invariants you name are invariant for the action of the larger group $GL_2 \times GL_2$, acting by $(g,h) \circ (m,n) \mapsto (g m g^{-1}, h n h^{-1})$. However, the polynomial I named is not invariant for this larger action. 
A: 
For  $G$ acting on $M_2(\mathbb{C})\times M_2(\mathbb{C})$ by $g\circ (m,n)=(gmg^{-1},gng^{-1})$, why is it that the ring of invariant polynomials under the group action is only $\mathbb{C}[\textrm{tr}(m),\det(m),\textrm{tr}(n),\det(n)]$? More precisely, why doesn't there exist other $G$-invariants using a combination of $m$ and $n$?

Your claim is very far from a true answer. In the book of Kraft and Procesi, Classical Invariant Theory, at page 21, they prove that all the invariants are generated by $\mathrm{tr}(A)$, $\mathrm{tr}(B)$, $\mathrm{tr}(A^2)$, $\mathrm{tr}(B^2)$ and $\mathrm{tr}(AB)$ which are algebraically independent. 
Remark. In the example of David Speyer the invariant element is $\mathrm{tr}(M)\mathrm{tr}(N)-\mathrm{tr}(MN)$.
