This is part of the implicit mapping theorem in Spivak's calculus on manifolds. We let $f:\Bbb{R}^n \times\Bbb{R}^m\to\Bbb{R}^m$, be continuously differentiable in an open set which contains a point $(a,b)$ such that $f(a,b) = 0$. Then, let $M$ be the $m\times m$ matrix, with the element $$(D_{n+j}f_i(a,b)),\enspace 1\leq i,j\leq m$$
as its row $i$ column $j$ element. For IMT, we need the determinant of this matrix to be zero. If we have a function $f(x,y) = x^2 + y^2 -1$, then isn't this matrix just $$D_y f(a,b)$$
the derivative with respect to $y$? ( $y$ can also be $x_2$ or the variable corresponding to the index $2$ ) So in general, this matrix contains the partials of $f$ with respect those variables that we wish to solve in terms of the variables excluded from this matrix ( in the example, this excluded variable is $x$ ). But I'm unsure about this, for if $f$ is a function of $(x_1,...,x_n,y_1,...,y_m)$, and if we would now like to solve $(y_1,...,y_m)$ in terms of $(x_1,...,x_n)$, then our matrix would look like $$(D_{n+1},D_{n+2},...,D_{n+m})$$
assuming $f$ is real valued. Then I suppose that $D_{n+1}=D_{y_1}$? Meaning we "jump" one variable to the right of $x_n$, and so on. Is this the right interpretation of $M$?
Your understanding is correct. You have a function $f : \mathbb R^n \times \mathbb R^m = \mathbb R^{n+m} \to \mathbb R^m$. It has $m$ coordinate functions $f_i : \mathbb R^{n+m} \to \mathbb R$ each of which depends on $n+m$ real variables. Thus $D_k f_i$ denotes the partial derivative of $f_i$ with respect to the $k$-th variable.
Let us neutrally write $f_i(z_1,\ldots,z_{n+m})$. Then $D_k f_i = \dfrac{\partial f_i}{\partial z_k}$. If we substitute $(z_1,\ldots,z_{n+m}) = (x_1,\ldots,x_n,y_1,\ldots,y_m)$, i.e. $z_k = x_k$ for $k \le n$ and $z_k = y_{n-k}$ for $k > n$, we get that what you write in your question: We have $D_{n+j}f_i = \dfrac{\partial f_i}{\partial y_j}$.