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UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.

First, we notice that $k^{R(i)} = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity: $$\sum_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$ by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+_{m-i} = (-1)^{m-i} B^-_{m-i}$, we rewrite the above identity as $$(\star)\qquad \sum_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$

Proof. Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being a nonnegative integer. We notice that $$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$ while $$B^-_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}_{x+1}(\frac1z),$$ where ${\cal B}_{x+1}(t)$ is the $(x+1)$-st Bernoulli polynomial and $[z^n]$ is the operator extracting the coefficient of $z^n$.

It follows that the l.h.s. of $(\star)$ equals $$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}_{x+1}(\frac1z).$$

Using Lagrange–Bürmann formula, we conclude that $$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}_{x+1}(\frac1z-1).$$ Then, by the properties of Bernoulli polynomials, $${\cal B}_{x+1}(\frac1z-1) = {\cal B}_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$

So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to $$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$ QED

UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.

First, we notice that $k^{R(i)} = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}=\binom{2i}{i}\cdot \frac{i!}{2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity: $$\sum_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$ by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+_{m-i} = (-1)^{m-i} B^-_{m-i}$, we rewrite the above identity as $$(\star)\qquad \sum_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$

Proof. Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being a nonnegative integer. We notice that $$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$ while $$B^-_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}_{x+1}(\frac1z),$$ where ${\cal B}_{x+1}(t)$ is the $(x+1)$-st Bernoulli polynomial and $[z^n]$ is the operator extracting the coefficient of $z^n$.

It follows that the l.h.s. of $(\star)$ equals $$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}_{x+1}(\frac1z).$$

Using Lagrange–Bürmann formula, we conclude that $$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}_{x+1}(\frac1z-1).$$ Then, by the properties of Bernoulli polynomials, $${\cal B}_{x+1}(\frac1z-1) = {\cal B}_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$

So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to $$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$ QED

UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.

First, we notice that $k^R(i) = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity: $$\sum_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$ by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+_{m-i} = (-1)^{m-i} B^-_{m-i}$, we get rewrite the above identity as $$(\star)\qquad \sum_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$

Proof. Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being a nonnegative integer. We notice that $$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$ while $$B^-_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}_{x+1}(\frac1z),$$ where ${\cal B}_{x+1}(t)$ is the $(x+1)$-st Bernoulli polynomial and $[z^n]$ is the operator extracting the coefficient of $z^n$.

It follows that the l.h.s. of $(\star)$ equals $$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}_{x+1}(\frac1z).$$

Using Lagrange–Bürmann formula, we conclude that $$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}_{x+1}(\frac1z-1).$$ Then, by the properties of Bernoulli polynomials, $${\cal B}_{x+1}(\frac1z-1) = {\cal B}_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$

So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to $$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$ QED

UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.

First, we notice that $k^{R(i)} = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity: $$\sum_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$ by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+_{m-i} = (-1)^{m-i} B^-_{m-i}$, we rewrite the above identity as $$(\star)\qquad \sum_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$

Proof. Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being a nonnegative integer. We notice that $$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$ while $$B^-_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}_{x+1}(\frac1z),$$ where ${\cal B}_{x+1}(t)$ is the $(x+1)$-st Bernoulli polynomial and $[z^n]$ is the operator extracting the coefficient of $z^n$.

It follows that the l.h.s. of $(\star)$ equals $$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}_{x+1}(\frac1z).$$

Using Lagrange–Bürmann formula, we conclude that $$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}_{x+1}(\frac1z-1).$$ Then, by the properties of Bernoulli polynomials, $${\cal B}_{x+1}(\frac1z-1) = {\cal B}_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$

So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to $$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$ QED

UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.

First, we notice that $k^R(i) = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity: $$\sum_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$ by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+_{m-i} = (-1)^{m-i} B^-_{m-i}$, we get rewrite the above identity as $$(\star)\qquad \sum_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$

Proof. Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being nonnegative integer. We notice that $$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$ while $$B^-_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}_{x+1}(\frac1z),$$ where ${\cal B}_{x+1}(t)$ is the $(x+1)$-st Bernoulli polynomial and $[z^n]$ is the operator extracting the coefficient of $z^n$.

It follows that the l.h.s. of $(\star)$ equals $$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}_{x+1}(\frac1z).$$

Using Lagrange–Bürmann formula, we conclude that $$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}_{x+1}(\frac1z-1).$$ Then, by the properties of Bernoulli polynomials, $${\cal B}_{x+1}(\frac1z-1) = {\cal B}_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$

So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to $$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$ QED

UPDATE. My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.

First, we notice that $k^R(i) = (k/2+i-1)_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity: $$\sum_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$ by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+_{m-i} = (-1)^{m-i} B^-_{m-i}$, we get rewrite the above identity as $$(\star)\qquad \sum_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$

Proof. Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being a nonnegative integer. We notice that $$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$ while $$B^-_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}_{x+1}(\frac1z),$$ where ${\cal B}_{x+1}(t)$ is the $(x+1)$-st Bernoulli polynomial and $[z^n]$ is the operator extracting the coefficient of $z^n$.

It follows that the l.h.s. of $(\star)$ equals $$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}_{x+1}(\frac1z).$$

Using Lagrange–Bürmann formula, we conclude that $$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}_{x+1}(\frac1z-1).$$ Then, by the properties of Bernoulli polynomials, $${\cal B}_{x+1}(\frac1z-1) = {\cal B}_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$

So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to $$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$ QED