Here goes the proof of the general fact. At first, we rewrite the product (ignoring the sign) as $\binom{r}{j}\cdot \binom{r+n-j-1}{n-j}$.

Denote $r=2^tR$ for odd $R$ and non-negative integer $t$.
If $\binom{r}{j}\cdot \binom{r+n-j-1}{n-j}$ is odd, then both $\frac{r}j\binom{r-1}{j-1}=\binom{r}j$ and $\frac{r}{n-j}\cdot \binom{r+n-j-1}{n-j-1}$ are odd. It implies that $2^{t}$ should divide both $j$ and $n-j$, therefore it divides $n$. It already proves the first case (in my notations it claims that $W(r,n)=0$ unless $2^t$ divides $n$).

If $n$ is also divisible by $2^t$, so are $j$ and $n-j$, and denoting $j=2^t J$, $n=2^t N$ we use the well-known congruence
$$\binom{2^t A}{2^t B}=[x^{2^t B}](1+x)^{2^t A}\equiv [x^{2^t B}](1+x^{2^t})^A\equiv [x^B](1+x)^A=\binom{A}B \pmod 2$$
for $A=R, B=J$ and $A=-R,B=N-J$ that yields $W(r,n)=W(R,N)$.

So we reduce to the case when $r$ is odd. Now we use Lucas' theorem that $\binom{a}b$ is odd if and only if $P(b)\subset P(a)$ where $P(a)$ is the set of powers of 2 in the binary expansion of $a$. Another equivalent reformulation is $P(a)\cap P(b-a)=\emptyset$.

In our situation it says that $\binom{r}{j} \binom{r+n-j-1}{n-j}$ is odd if and only if $P(j)\subset P(r)$, $P(n-j)\cap P(r-1)=\emptyset$.

Note that since $r$ is odd, we have $P(r)=P(r-1)\sqcup \{2^0\}$.

Now consider two cases.

1) $j$ is even. Then the sets $P(j)\subset P(r-1)$ and $P(n-j)$ are disjoint, thus their union is $P(n)$ and so we have unique possibility $P(j)=P(r-1)\cap P(n)$, $P(n-j)=P(n)\setminus P(r-1)$.

2) $j$ is odd. Then $j-1$ is even and $P(j)\subset P(r)$ reads as $P(j-1)\subset P(r-1)$. Analogously, we get the unique solution $P(j-1)=P(r-1)\cap P(n-1)$, $P(n-j)=P(n-1)\setminus P(r-1)$.