What is the dimension of a vector-valued function space?

Question

For example, if we have a vector-valued function space

$$ S = [V^{k+1,k}, V^{k,k+1}]^T $$ where the component space in two coordinates is $$ V^{m,n} := \{v: \mathbb{R}^2 \to \mathbb{R} \mid v = \sum_{i=0}^m \sum_{j=0}^n c_{i,j} x_1^i x_2^j,~~ x_1 \in \mathbb{R}, x_2 \in \mathbb{R} \} $$

1. What is the dimension of $V^{m,n}$? I think there are $(m+1)(n+1)-1$ basis functions spanned the space, where the constant $c_{0,0} x_1^0 x_2^0$ should be excluded, so $\operatorname{dim}(V^{m,n}) = (m+1)(n+1)-1$. Is that correct?

2. What is the dimension of $S$? Shall we add up the dimension of each component space, or take square? If my answer to Q1 is correct, should the $\operatorname{dim}(S)$ be $2((k+2)(k+1)-1)$ or $((k+2)(k+1)-1)^2$?