How to show independence and uniform distribution of hash codes from k-wise independent hash functions?
Throughout, we assume that $|D| \geq k$.
Suppose that $H$ satisfies, for all distinct $x_1,\dots,x_k \in D$ and all $y_1,\ldots,y_k \in R$,
$$
\Pr_{h \in H}[h(x_1)=y_1,\ldots,h(x_k)=y_k] = \frac{1}{|R|^k}.
$$
Now let $x \in D$ be arbitrary. Since $|D| \geq k$, we can find $x_2,\ldots,x_k \in D$ such that $x,x_2,\ldots,x_k$ are distinct. For each $y \in R$,
$$
\Pr_{h \in H}[h(x)=y] =
\sum_{y_2,\ldots,y_k \in R} \Pr_{h \in H}[h(x)=y,h(x_2)=y_2,\ldots,h(x_k)=y_k] = \\
\sum_{y_2,\ldots,y_k \in R} \frac{1}{|R|^k} = \frac{|R|^{k-1}}{|R|^k} = \frac{1}{|R|}.
$$
This proves (i). To see (ii), let $x_1,\ldots,x_k \in D$ be distinct. Then for all $y_1,\ldots,y_k$,
$$
\Pr_{h \in H}[h(x_1)=y_1,\ldots,h(x_k)=y_k] = \frac{1}{|R|^k} = \prod_{i=1}^k \frac{1}{|R|} = \prod_{i=1}^k \Pr_{h \in H}[h(x_i) = y_i].
$$
In the other direction, suppose that (i) and (ii) hold. Then for all distinct $x_1,\ldots,x_k \in D$ and $y_1,\ldots,y_k \in R$,
$$
\Pr_{h \in H}[h(x_1)=y_1,\ldots,h(x_k)=y_k] = \prod_{i=1}^k \Pr_{h \in H}[h(x_i) = y_i] = \prod_{i=1}^k \frac{1}{|R|} = \frac{1}{|R|^k} .
$$