I have observed that the following identity seens to hold:
Let $k \geq 2$ and $0 \leq p, q \leq k+1$. Then $$\sum_{m = 0}^{k+1-q} \begin{pmatrix} -\frac12 + \lceil \frac{m+q}{2} \rceil \\ m \end{pmatrix} \begin{pmatrix} -\frac12 - \lceil \frac{m+q}{2} \rceil \\ p + q - m \end{pmatrix} = \begin{cases} 0 & q \neq p \\ 1 & q = p \end{cases}$$
I could not prove it yet. I have attempted to get it into a form where I can apply a kind of (alternating) Chu-Vandermonde identity, unsuccessfully. Binomial coefficients with an offset by $\frac12$ may be written in terms of double factorials. I have tried to rewrite the whole thing in terms of double factorial binomial coefficients (denoted by two brackets) but this also did not lead anywhere yet. Perhaps one should split into even and odd terms and look for identities summing over only even or odd $m$.
The calculations below may be useful. Bonne chance! It would be great if someone can find a proof. \begin{align*} &\qquad\;\;\; \sum_{m = 0}^{k+1-q} \begin{pmatrix} -\frac12 + \lceil \frac{m+q}{2} \rceil \\ m \end{pmatrix} \begin{pmatrix} -\frac12 - \lfloor \frac{m+q}{2} \rfloor \\ p + q - m \end{pmatrix} \\ &= \sum_{m = 0}^{(p+q) \land (k+1-q)} \frac{\Gamma(\frac12 + \lceil \frac{m+q}{2} \rceil)}{m! \, \Gamma(\frac12 + \lceil \frac{m+q}{2} \rceil - m)} \frac{\Gamma(\frac12 - \lfloor \frac{m+q}{2} \rfloor)}{(p + q - m)! \, \Gamma(\frac12 - \lfloor \frac{m+q}{2} \rfloor + m - p - q)} \\ &= \sum_{m = 0}^{(p+q) \land (k+1-q)} \frac{1}{m! \, (p + q - m)!} \frac{(2 (\lceil \frac{m+q}{2} \rceil) - 1)!!}{2^{\lceil \frac{m+q}{2} \rceil}} \frac{(-2)^{\lfloor \frac{m+q}{2} \rfloor}}{(2(\lfloor \frac{m+q}{2} \rfloor) - 1)!!} \\ &\qquad\;\;\; \cdot \frac{(2 (\lfloor \frac{m+q}{2} \rfloor - m + p + q) - 1)!!}{(-2)^{\lfloor \frac{m+q}{2} \rfloor - m + p + q}} \begin{cases} \frac{2^{\lceil \frac{m+q}{2} \rceil - m}}{(2 (\lceil \frac{m+q}{2} \rceil - m) - 1)!!} &, \lceil \frac{m+q}{2} \rceil - m \geq 0 \\ \frac{(2 (m - \lceil \frac{m+q}{2} \rceil) - 1)!!}{(-2)^{m - \lceil \frac{m+q}{2} \rceil}} &, \lceil \frac{m+q}{2} \rfloor - m < 0 \\ \end{cases} \\ &= \frac{(-1)^{p+q}}{2^{p+q} (p+q)!}\sum_{m = 0}^{(p+q) \land (k+1-q)} \begin{pmatrix} p+q \\ m \end{pmatrix} \begin{cases} 1 &, m+q \text{ even} \\ m+q &, m+q \text{ odd} \end{cases} \\ &\qquad\;\;\; \cdot \begin{cases} (-1)^m \frac{(2 (\lfloor \frac{m+q}{2} \rfloor - m + p + q) - 1)!!}{(2 (\lceil \frac{m+q}{2} \rceil - m) - 1)!!} &, \lceil \frac{m+q}{2} \rceil - m \geq 0 \\ (-1)^{\lceil \frac{m+q}{2} \rceil} \frac{(2 (\lfloor \frac{m+q}{2} \rfloor - m + p + q) - 1)!! \, (2 (m - \lceil \frac{m+q}{2} \rceil) - 1)!!}{1} &, \lceil \frac{m+q}{2} \rceil - m < 0 \\ \end{cases} \\ &= \frac{(-1)^{p+q}}{2^{p+q} (p+q)!}\sum_{m = 0}^{(p+q) \land (k+1-q)} \begin{pmatrix} p+q \\ m \end{pmatrix} \\ &\qquad\;\;\; \cdot \begin{cases} (-1)^m \frac{(2 (\frac{m+q}{2} - m + p + q) - 1)!!}{(2 (\frac{m+q}{2} - m) - 1)!!} &, m+q \text{ even}, \frac{m+q}{2} - m \geq 0 \\ (-1)^m (m + q) \frac{(2 (\frac{m+q-1}{2} - m + p + q) - 1)!!}{(2 (\frac{m+q+1}{2} - m) - 1)!!} &, m+q \text{ odd}, \frac{m+q+1}{2} - m \geq 0 \\ (-1)^{\frac{m+q}{2}} (2 (\frac{m+q}{2} - m + p + q) - 1)!! \, (2 (m - \frac{m+q}{2}) - 1)!! &, m+q \text{ even}, \frac{m+q}{2} - m < 0 \\ (-1)^{\frac{m+q+1}{2}} (m + q) (2 (\frac{m+q-1}{2} - m + p + q) - 1)!! \, (2 (m - \frac{m+q+1}{2}) - 1)!! &, m+q \text{ odd}, \frac{m+q+1}{2} - m < 0 \\ \end{cases} \\ &= \frac{(-1)^{p+q}}{2^{p+q} (p+q)!}\sum_{m = 0}^{(p+q) \land (k+1-q)} \begin{pmatrix} p+q \\ m \end{pmatrix} \\ &\qquad\;\;\; \cdot \begin{cases} (-1)^m \frac{(q - m - 1+ 2 (p + q))!!}{(q - m - 1)!!} &, m+q \text{ even}, \frac{m+q}{2} - m \geq 0 \\ (-1)^m (m + q) \frac{(q - m - 2 + 2 (p + q))!!}{(q - m)!!} &, m+q \text{ odd}, \frac{m+q+1}{2} - m \geq 0 \\ (-1)^{\frac{m+q}{2}} (q - m - 1 + 2 (p + q))!! \, (m - q - 1)!! &, m+q \text{ even}, \frac{m+q}{2} - m < 0 \\ (-1)^{\frac{m+q+1}{2}} (m + q) (q - m - 2 + 2 (p + q))!! \, (m - q - 2)!! &, m+q \text{ odd}, \frac{m+q+1}{2} - m < 0 \\ \end{cases} \\ &= \frac{(-1)^{p+q}}{2^{p+q} (p+q)!}\sum_{m = 0}^{(p+q) \land (k+1-q)} \begin{pmatrix} p+q \\ m \end{pmatrix} \\ &\qquad\;\;\; \cdot \begin{cases} (-1)^m (2 (p + q))!! \begin{pmatrix}\begin{pmatrix} q - m - 1 + 2 (p + q) \\ 2 (p + q) \end{pmatrix}\end{pmatrix} &, m+q \text{ even}, q \geq m \\ (-1)^m (m + q) (2 (p + q - 1))!! \begin{pmatrix}\begin{pmatrix} q - m - 2 + 2 (p + q) \\ 2 (p + q - 1) \end{pmatrix}\end{pmatrix} &, m+q \text{ odd}, q + 1 \geq m \\ (-1)^{\frac{m+q}{2}} (2 (p + q - 1))!! \begin{pmatrix} \begin{pmatrix} 2 (p + q - 1) \\ m - q - 1 \end{pmatrix} \end{pmatrix}^{-1} &, m+q \text{ even}, q < m \\ (-1)^{\frac{m+q+1}{2}} (m + q) (2 (p + q - 2))!! \begin{pmatrix} \begin{pmatrix} 2 (p + q - 2) \\ m - q - 2 \end{pmatrix} \end{pmatrix}^{-1} &, m+q \text{ odd}, q + 1 < m \\ \end{cases} \end{align*}