I. In the first case, you can use the fact that if $$\cdots\to P_n\to P_{n-1}\to\cdots\to P_1\to P_0$$ is a projective resolution of $N$, then every Yoneda $n$-extension of $M$ by $N$ can be represented an extension of the form $$ 0\to M\to E\to P_{n-2}\to P_{n-3}\to\cdots \to P_1\to P_0\to N\to 0$$ and where $E$ is a module which is constructed as a pushout of a diagram of the form $$M\leftarrow P_n\rightarrow P_1$$ Using this, you get a sensible set of representatives of $n$-extensions, which you can probably make into a scheme. You next want to quotient by equivalence---I do not see immediately how that'll work.
II. For the second case, and if you are considering finite dimensional modules over a finitely generated algebra $A$, you can construct an analogue of the representation variety $\mathrm{Rep}_d(A)$ for filtered modules with specific subquotients. For example, suppose you want a variety of modules $M$ of total dimension $d$ with a filtration $0=F_0\subseteq F_1\subseteq F_2\subseteq F_3=M$ such that $F_1/F_0\cong N_1$, $F_2/F_2\cong N_2$ and $F_3/F_2\cong N_3$. Up to isomorphism, you can suppose that $M=k^d$, and that the $F_i$ are a standard partial flag (so that $F_i$ is the subspace of $k^d$ of vectors whose last $d-\dim F_i$ coordinates vanish)
The action of $M$ is then completely given by $n$ $d$-by-$d$ matrices, where $n$ is the size of a generating set of $A$, and the fact that $M$ is an actual module, that chosen filtration is a module filtration, and that the subquotients are what they should be can be expressed in terms of polynomial equations involving the coefficients of those $n$ matrices.
This determines a scheme, whose points are $A$-module structures on $k^d$ which satisfy the desired conditions, and which contain representatives of all isoclasses of modules satisfying those conditions. Of course, the points of this scheme are not in correspondence with isoclasses: to do that, you need to pass to the quotient by the appropriate change-of-basis group (but that will kill the scheme structure, I guess...)