I have questions about [Risch 1979] (see the citation and reference below).
1.) What are the factors of the exterior product of the exterior differential algebra in the citation below?
I already know the terms exterior algebra and differential algebra.
2.) Is $\Omega^i(\mathfrak{D})$ the set of the $i$-forms on $\mathfrak{D}$?
"First some notation. We will work with fields $\mathfrak{D}$, $\mathfrak{E}$,... that are always finitely generated over some algebraically closed field $\textit{K}$ of characteristic $0$ (...). If $x_1,...,x_t$, is a transcendence basis of $\mathfrak{D}/\textit{K}$, then all $\textit{K}$ derivations of $\mathfrak{D}$ are of the form $\sum_{i=1}^t\ g_i\partial/\partial x_i$, $g_i\in \mathfrak{D}$. $\Omega(\mathfrak{D})$ = (the exterior differential algebra formed from $\mathfrak{D}$) = $\sum_0^t\Omega^i(\mathfrak{D})$, with $\Omega^0(\mathfrak{D})=\mathfrak{D}$, $\Omega^1(\mathfrak{D})=\{\sum_{i=1}^tg_idx_i\colon g_i\in\mathfrak{D}, dx_i(\partial/\partial x_j)=\delta_j^i\}$ = dual space to the derivations of $\mathfrak{D}/\textit{K}$."
Disclaimer: My exposure to these notions is rather small. Take everything with a grain of salt.
Short answer: Most likely, what's dubbed an exterior differential algebra here is actually the exterior algebra over the so called “Module of Kähler differentials of $\mathfrak D$ over $K$”, written $\Omega_{\mathfrak D /K}$, which in turn should have the structure of a differential (graded) algebra in a canonical way.
Long answer: It appears that Eisenbud[1] provides the necessary basics here:
The context is a commutative ring $R$ and an $R$-Algebra $S$. Given these and an $S$-Module $M$ (in our case, $R=K$ and $S=M=\mathfrak D$), we can define $\operatorname{Der}_R(S, M)$ to be all the $R$-linear maps $S\to M$ satisfying the Leibniz rule. This has evidently the structure of an $S$-module by the usual pointwise defined action.
As it turns out, there is a universal object describing these: Every such derivation corresponds uniquely to an $S$-linear map $\Omega_{S/R}\to M$, where $\Omega_{S/R}$ is the $S$-Module generated by symbols $\{d(f)\mid f\in S\}$ subject to the relations $d(fg) = fd(g) + gd(f)$ for all $f, g\in S$ and $d(\lambda f + g) f \lambda d(f) + d(g)$ for all $f, g\in S$ and $\lambda\in R$. Even stronger, we have $$ \operatorname{Der}_R(S, M)\simeq \operatorname{Hom}_S(\Omega_{S/R}, M), $$ where the underlying isomorphism is $S$-linear and natural in $M$.
Now if we head to our case, setting $R = K, S = M = \mathfrak D$, the isomorphism gives us $$ \operatorname{Der}_K(\mathfrak D) := \operatorname{Der}_K(\mathfrak D, \mathfrak D)\simeq \operatorname{Hom}_{\mathfrak D}(\Omega_{\mathfrak D/K}, \mathfrak D) =: \Omega_{\mathfrak D/K}{}^\ast, $$ and assuming reflexivity, interpreting $\Omega(\mathfrak D)$ as the usual exterior algebra of $\mathfrak D$-modules $\bigwedge \Omega_{\mathfrak D/K}$ should work.
The reflexivity should hold due to the remark at the beginning, where $\operatorname{Der}_K(\mathfrak D)$ is identified as a free $\mathfrak D$-module over the derivations $\partial/\partial x_i$, and free modules are obviously reflexive, since we can provide an explicit dual basis (here denoted by $dx_i$).
I don't know a proof of this off the top of my head, but I recommend looking at the simple case $K=\mathbb Q$ and one transcendental element. Intuitively, by the Leibniz rule, every differential should be given by its value at its transcendental element, and transcendental elements should be assignable to every value in the large field. But this is by no means formal.
This naked exterior algebra can now be equipped with a differential structure by means of the map $$ d^i (g d(f_1) \wedge d(f_2) \wedge \ldots \wedge d(f_i)) := dg \wedge d(f_1) \wedge d(f_2) \wedge \ldots \wedge d(f_i), $$ Turning it into a differential graded algebra.
[1] Chapter 16, “Modules of Differentials” in David Eisenbud - Commutative Algebra – with a View Toward Algebraic Geometry, 1995.