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The green result can be used to prove the red one, yes.

Start with the fact that $\binom{m}{k} = \binom{m}{m - k}$. If this isn't obvious to you, you can derive it either by looking at the usual formula for the binomial, or by considering that choosing the members of a subset is equivalent to choosing elements *excluded* from the subset.

You can use that fact to see that $\sum_{k=0}^n \binom{2n + 1}{k} = \sum_{k=0}^n \binom{2n + 1}{2n + 1 - k}$. Considering the set of numbers $\\{2n + 1 - k \mid k \in \\{ 0,\dots,n \\}\\}$, you can see that this set is the same as $\\{n + 1,\dots, 2n + 1\\}$, so another way to write $\sum_{k=0}^n \binom{2n + 1}{2n + 1 - k}$ is $\sum_{k=n + 1}^{2n + 1} \binom{2n + 1}{k}$.

Then, of course, $\sum_{k=0}^n \binom{2n + 1}{k} + \sum_{k=n + 1}^{2n + 1} \binom{2n + 1}{k} = \sum_{k=0}^{2n + 1} \binom{2n + 1}{k}$, which by the green equation is equal to $2^{2n + 1}$. But from the previous paragraph, we know the summands on the left are equal to each other, so they must each be equal to half of the result, or $2^{2n}$. This proves the red equation. (You need a little more to go the other way; using this line of reasoning starting with the red equation would only prove that the green equation holds for odd $n$.)