Question 1: The general construction of the Schur module (see for instance Fulton's book on Young tableaux, Chapter 8; the example $S_{2,1}(V)$ appears in the introduction of Part II) yields
$S_{2,1}(V) = \mathrm{coker}(\omega: \Lambda^3(V) \to \Lambda^2(V) \otimes V),$
where $\omega$ is defined by $$\omega(a \wedge b \wedge c) = (b \wedge c) \otimes a - (a \wedge c) \otimes b + (a \wedge b) \otimes c,$$ or equivalently by the more symmetric definition $$\omega(a \wedge b \wedge c) = (a \wedge b) \otimes c + (b \wedge c) \otimes a + (c \wedge a) \otimes b.$$ Notice that $\omega$ is part of the comultiplication of the graded Hopf algebra $\Lambda^*(V)$. You can also write $S_{2,1}(V)$ as a kernel, but let me work with this cokernel description.
Notice that $\omega$ has a left inverse $s : \Lambda^2(V) \otimes V \to \Lambda^3(V)$ given by mapping $(a \wedge b) \otimes c$ to $\frac{1}{3} (a \wedge b \wedge c)$. In particular, $\omega$ is injective. Hence, $$\dim(S_{2,1}(V)) = \binom{n}{2} \cdot n - \binom{n}{3} = \frac{(n+1)n(n-1)}{3}.$$
We can also write down a basis: If $e_1,\dotsc,e_m$ is a basis of $V$, then the images of the vectors $(e_i \wedge e_j) \otimes e_k$ with $i<j$ and $i \leq k$ form a basis of $S_{2,1}(V)$. For this one checks (a) that they generate $S_{2,1}(V)$, (b) that their number is $\dim(S_{2,1}(V))$.
Here is a sketch of a completely different approach: Using the decomposition $$\Lambda^n(V \oplus W) = \bigoplus_{p=0}^{n} \Lambda^p(V) \otimes \Lambda^{n-p}(W),$$ one finds a decomposition $$S_{2,1}(V \oplus W) \cong S_{2,1}(V) \oplus S_{2,1}(W) \oplus (V^{\otimes 2} \oplus W) \oplus (V \otimes W^{\otimes 2}).$$ In particular, we have $$S_{2,1}(V \oplus \mathbb{K}) \cong S_{2,1}(V) \oplus V^{\otimes 2} \oplus V.$$ From here, it is easy to determine the dimension recursively.