Let me discuss the graded case, that is the ring is the polynomial ring and the matrices have general homogeneous entries of degree 1$1$. The local version should follows by taking "lowest order part" as you do in example 1$1$.
Consider fistfirst of the case of the polynomial ring S$S$ in variables x_{ij} 's$x_{ij}$ and y_{ij} 's$y_{ij}$ and matrices X=(x_{ij}) of$X=(x_{ij})$ of size axb$a\times b$ and Y=(y_{ij})$Y=(y_{ij})$ of size cxd$c\times d$ with a\leq b$a\leq b$ and c\leq d$c\leq d$. Then take the ideal I$I$ of minors of size a$a$ of X$X$ and the ideal J$J$ of minors of size c$c$ of Y$Y$.
Set A=S/I+J$A=S/I+J$.
The minimal free resolution of A$A$ is obtained by taking the tensor product of the resolution of S/I$S/I$ with that of S/J$S/J$ because the ideals are are generated by polynmialspolynomials in different varaiblesvariables. We may use this fact, in combination with the fact that I$I$ and J$J$ are resolved by the Eagon-Northcot complex, toto compute the Castelnuovo-Mumford reg(A)regularity $\operatorname{reg}(A)$ of A$A$, its projective dimension and its dimension. WeWe have:
reg(A)=reg(S/I)+reg(S/J)=(a-1)+(b-1).$\operatorname{reg}(A)=\operatorname{reg}(S/I)+\operatorname{reg}(S/J)=(a-1)+(b-1)$
dim(A) = ab+cd-(b-a+1)-(d-c+1)$\dim(A) = ab+cd-(b-a+1)-(d-c+1)$
and A$A$ is Cohen-Macaulay.
Now we specialize generically the x_{ij}$x_{ij}$'s and the y_{ij}$y_{ij}$'s generically to linearlinear forms inin variables z_1,..,z_n$z_1,..,z_n$. The ring you want to udenrstandunderstand gets identified with A/L $A/L$ where L$L$ is generated by ab+cd-n$ab+cd-n$ general linear forms in the x_{ij}$x_{ij}$ and y_{ij}$y_{ij}$.
Now if n\leq (b-a+1)+(d-c+1)$n\leq (b-a+1)+(d-c+1)$ then ab+cd-n\geq ab+cd-(b-a+1)-(d-c+1)$ab+cd-n\geq ab+cd-(b-a+1)-(d-c+1)$ and ab+cd-(b-a+1)-(d-c+1)$ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating L$L$ form a maximal regular sequence in A$A$. Let U$U$ be the ideal generated by ab+cd-(b-a+1)-(d-c+1) of$ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating L$L$.
Then reg(A/U)=reg(A)=(a-1)+(b-1)$\operatorname{reg}(A/U)=\operatorname{reg}(A)=(a-1)+(b-1)$ and dim(A/U)=0$\dim(A/U)=0$. The regularity for a 0$0$-dimensional module M is$M$ is the largest index i$i$ such that M_i\neq 0$M_i\neq 0$. It follows that the (A/U)_i=0$(A/U)_i=0$ for i>(a-1)+(b-1)$i>(a-1)+(b-1)$. And the same is true for A/L $A/L$ (becasuebecause it is a quotinetquotient of A/U$A/U$). Hence we have (A/L)_i=0$(A/L)_i=0$ for i>(a-1)+(b-1)$i>(a-1)+(b-1)$. On the size of the z's$z$'s it says that the ideal (z_1,..,z_n)$(z_1,..,z_n)$ to the power (a-1)+(b-1)+1$(a-1)+(b-1)+1$ is contained in ideal of definitondefinition, which is exactly what you wanted to prove.