The Vandermonde's Identity tells us:
$\displaystyle \large {m+n \choose r}=\sum_{k=0}^r{m \choose k}{n \choose r-k}$
However, I want to find out the closed form or the simplest form of the expression:
$\displaystyle \large \sum_{r=1}^{p}\sum_{k=0}^r{m \choose k}{n \choose r-k}$, where $p$ is a natural number.
But I only want to include the terms with odd $k$ to be included in the inner sum (the $\displaystyle \large \sum_{k=0}^r{m \choose k}{n \choose r-k}$ one)
I had previously posted a similar question before, but I couldn't get a simplification for that case. However, with a change in the nature of the sum I believe will render it to have a simplification, at least this time.
Idea. The original vandermonde identity is a consequence of
$$ (1+x)^m (1+x) ^n = (1+x) ^{m+n}$$
(look at the coefficient of $x^r$ !!). Now we do this trick: if we want only odd $k$ to appear, we should consider $((1+x) ^m - (1-x) ^m ) /2$ (why? This is an instructive exercise). In this way we get,
$$ ((1+x) ^m - (1-x) ^m )(1+x) ^n /2 = ((1+x) ^{m+n} - (1-x)^m (1+x) ^n )/2$$
Unfortunately, I don't think this will have a closed form. Nevertheless, for $p=m+n $ in which we sum all terms we have a lucky strike. Summing up all the coefficients of a polynonial is like calculating it in 1, so that the result is $2^{m+n}$. Since you wanted to exclude $r=$0 which contributed with +1, you'll get $2^{m+n}-1$.
These techniques are based on generating functions, which I encourage you to deepen since they are a wonderful subject. Hard equalities like Vandermonde comes from very easy algebraic manipulations.