Looking at (3.20) in Gould's tables, I have an idea of what might be his intended proof of the second identity. At the end of this answer I've indicated a bijective proof, using the same method as Gjergji Zaimi's answer. (This was a comment, but on rereading it, it didn't seem clear to me, so I'e expanded it below.)

It suffices to show that

$$\tag{$\star$}\sum_{j=0}^k \binom{2n}{2j} \binom{n-j}{k-j} = \frac{2^{2k}}{(2k)!}\prod_{j=0}^{k-1}(n^2-j^2)$$

since the right-hand side is easily seen to be $\frac{2^{2k}}{2k}n\binom{n+k-1}{2k-1} = 2^{2k}\binom{n+k}{2k}\frac{n}{n+k}$.

Let $L(n)$ denote the left-hand side and $R(n)$ denote the right-hand side in ($\star$). Each is a polynomial in $n$. Clearly $L(n)$ has roots for $n \in \{0,1,\ldots, k-1\}$. When $n=k$ we get $\sum_{j=0}^k \binom{2k}{2j} = 2^{2k-1}$. In all cases this agrees with $R(n)$. Evaluating $L(n)$ at $n=-m$ we get

$$ \sum_{j=0}^k \binom{-2m}{2j} \binom{-m-j}{k-j} = \sum_{j=0}^k \binom{2j+2m-1}{2m-1} \binom{k+m-1}{k-j}(-1)^{k-j} $$

When $m \ge 2$ we can extend the sum so that the bottom limit is $-(m-1)$ since each binomial coefficient $\binom{2j+2m-1}{2m-1}$ vanishes for $j \in \{-1,\ldots, -(m-1)\}$. The right-hand side above is now the degree $k+m-1$ difference operator, $\sum_{\ell=0}^{k+m-1} \binom{k+m-1}{\ell}(-1)^{\ell}$ applies to the polynomial $f(\ell) = \binom{2k + 2m - 1 - 2\ell}{2m-1}$ of degree $2m-1$. Hence $L(n)$ has roots for $n \in \{-1,\ldots,-(k-1)\}$.

When $m=k$ the difference operator gives $(2k-1)!$ times the leading coefficient of $f(-\ell)$, namely $2^{2k-1}/(2k-1)!$. Therefore $L(-k) = 2^{2k-1}$; we saw above that $R(k) = 2^{2k-1}$, and clearly $R$ is even, therefore $L(-k) = R(-k)$.

This shows that $L$ and $R$ are polynomials in $n$ of degree $2k$ agreeing for all $n \in \{-k,\ldots, k\}$. Hence they are equal.

*Remark.* This proof could be significantly shortened if there was some easy reason why $L(n)$ is an even function of $n$. (As it is, this is only clear by the end of the proof.) In Gould's tables, $L(n)$ appears by expanding
$\cos 2nx = \Re (\cos x + \mathrm{i} \sin x)^{2n} = \sum_{j} \binom{2n}{2j} (-1)^{j} \sin^{2j}\!x (1-\sin^2\!x)^{n-j}$ to get

$$ \cos 2nx = \sum_{k=0}^n (-1)^k \sin^{2k}\!x \Bigl( \sum_{j=0}^k \binom{2n}{2j} \binom{n-j}{k-j} \Bigr). $$

If the right-hand side could somehow be defined for negative $n$, then the obvious evenness of $\cos 2nx$ would give the result. But I cannot see a way to make this formal argument correct.

*Bijective proof.* Start with $2n$ boxes in a row and choose $2(n-j)$ to colour white, black, $\ldots$, white, black. Then choose $n-k$ of the $n-j$ white boxes to mark. Next delete every box following a marked white box (the final box is either black or uncoloured, so this can be done), and the first box (which might be marked), and identify white and black. The resulting configuration has $n+k-1$ boxes, of which either $n-k$ or $n-k+1$ are white and marked, and some of the remaining boxes are coloured (with an unknown colour).

Given a configuration of the first type, insert a new box after each marked (white) box, and a new first box. The colours can then be reconstructed, as in Zaimi's proof, by working from right to left: if one has just coloured a box white (resp. black), and the next box to the left is marked, then the inserted box must be coloured black (resp. uncoloured). For the second type, insert a new box after each marked box, and *two* boxes at the start, the first of which is marked white, and then reconstruct the colours as before.

Hence

$$ \sum_{j=0}^n \binom{2n}{2(n-j)}\binom{n-j}{k-j} = 2^{2k-1}\binom{n+k-1}{n-k} + 2^{2k}\binom{n+k-1}{n-k+1} $$

The right-hand side simplifies to

$$2^{2k}\binom{n+k}{n-k} \Bigl( \frac{k}{n+k} + \frac{n-k}{n+k} \Bigr) = 2^{2k} \binom{n+k}{2k} \frac{n}{n+k} $$

as required. The third identity can be proved very similarly.