Question: "Is Wd being treated as a variety, and is Wd(k) its k-rational points?"
Answer: If $V:=k\{e_1,..,e_n\}$ is a vector space over a field $k$ where you have chosen a basis $e_i$, you get a "dual basis" $x_i:=e_i^*$ for the dual $V^*$ where $x_i(e_j):=\delta_{ij}$. The "ring of polynomial functions" on $V$ is "by definition" the $k$-algebra $Sym_k^*(V^*) \cong k[x_1,..,x_n]$, which is the polynomial ring on the independent variables $x_i$. The "variety of $V$" is the affine scheme
$$\mathbb{V}(V^*):=Spec(Sym_k^*(V^*)).$$
We write $V^*$ in the above to notify that we take the symmetric algebra in $V^*$ to obtain the polynomial functions on $V$. This construction is coordinate-free and functorial in $V$ - we construct the ring of polynomial functions on $V$ intrinsically from $V$. When $V$ has an action of a group $G$, there is a canonical action on $V^*$ and $Sym_k^*(V^*)$. If you ignore the dual in this correspondence you end up making mistakes. This may seem like a "triviality" - a vector space and its dual have the same dimension, so why bother?
Example: Hence if you choose a basis $W_d:=k\{e_1,..,e_n\}$, the dual vector space $W_d^*$ has basis $x_i$, and the polynomial functions on $W_d$ is by definition the polynomial ring $k[x_1,..,x_n]$. Let $S:=Spec(k[x_i])$. With this definition it follows $W_d(k)$ is the set of $k$-rational points of $S$. This is the set of maximal ideals $\mathfrak{m}:=(x_1-a_1,..,x_n-a_1)$ with $a_i \in k$. Hence there is an equality of sets
$$W_d(k) \cong k^n.$$
Example: To keep track of duals of vector spacces is important when you are considering group actions. Then $V$ and $V^*$ are not isomorphic as representations in general.
Let $V:=k\{e_0,e_1\}$ and $V^*:=k\{x_0,x_1\}$ and let $S:=SL(V)$ and $G:=GL(V)$ act on $V$ and $V^*$. There is a canonical map
$$V\otimes_k V \rightarrow \wedge^2 V$$
giving an isomorphism $V \cong V^*\otimes \wedge^2 V$ as $G$-modules. The vector space $\wedge^2 V$ is a trivial $S$-module, hence the $S$-modules $V$ and $V^*$ are isomorphic. If $P:=\mathbb{P}(V^*)$ is the projective space on $V$ it follows $H^0(P, \mathcal{O}(d)) \cong Sym^d(V^*)$ and there is no isomorphism between $Sym^d(V)$ and $Sym^d(V^*)$ as $G$-modules. Hence you must take care when calculating global sections - if you do not take care about the choice of variables you will end up working with the wrong representation: $V$ instead of $V^*$.
