Expression: ${m}\choose{1}$${n-m}\choose{k-1}$ + ${m}\choose{3}$${n-m}\choose{k-3}$ + ... + ${m}\choose{k}$${n-m}\choose{0}$ (assuming k is odd)
We could write it as ${n}\choose{k}$ if the even terms were also present, but I am clueless as they are not.
I think there is no simple formula for the first expression, let alone the second one.
We have $$\sum_{j=0}^k \binom{m}{j}\binom{n-m}{k-j} x^j = [y^k]\ (1+xy)^m(1+y)^{n-m},$$ where $[y^k]$ is the operator taking coefficient of $y^k$. Then the sum of coefficients at odd $k$ equals halved difference of the above expression evaluated at $x=1$ and at $x=-1$, that is \begin{split} \sum_{j=0\atop j\text{ is odd}}^k \binom{m}{j}\binom{n-m}{k-j} &= [y^k]\ \frac{(1+y)^m(1+y)^{n-m} - (1-y)^m(1+y)^{n-m}}2 \\ &=\frac{1}{2}\binom{n}{k} - \frac{1}{2} \sum_{i=0}^k (-1)^i\binom{m}i\binom{n-m}{k-i}. \end{split} The last sum here simplifies is some cases -- e.g., if $n=2m$, then it's zero.
For the second expression, $$\sum_{j=0}^k \binom{m}{j} x^j\sum_{i=0}^{k-j}\binom{n-m}{i} = [y^k]\ (1+xy)^m\frac{(1+y)^{n-m}}{1-y}.$$ Hence, we similarly have \begin{split} \sum_{j=0\atop j\text{ is odd}}^k \binom{m}{j}\sum_{i=0}^{k-j}\binom{n-m}{i} &= \frac{1}{2} [y^k]\ \big(\frac{(1+y)^n}{1-y} - (1-y)^{m-1}(1+y)^{n-m}\big) \\ &= \frac{1}{2}\sum_{i=0}^k\big(\binom{n}{i} - (-1)^i\binom{m-1}i \binom{n-m}{k-i}\big). \end{split}