Suppose I have the following equality:
$$\sum_{k=0}^{n-a}\sum_{j=0}^{k}\binom{n}{k}\binom{k}{j}\frac{f(a,k)\cdot g(b,n-k)}{n!}=\sum_{k=0}^{n-a}\binom{n}{k}\binom{n-k}{a}\frac{z^k \cdot g(b,n-k-a)}{n!}$$ for multivariate functions $f,g,$ and with $a,b,z\in \mathbb{C}$. Considering we are talking about finite sums, I feel like I can just simply drop the outer summation, drop the first binomial number $\binom{n}{k}$ and then drop the $n!$ as well since they are common for all terms in each summand. Thus I could say that, more simply,
$$\sum_{j=0}^{k}\binom{k}{j}f(a,k),g(b,n-k)=\binom{n-k}{a}z^kg(b,n-k-a)$$
I don't see the problem in this. I think because the $n$ dropped out I'll need to address the fact that $n>k$ in my final solution, but is this simplification okay?
No. Why would it be okay? If $1+2+3+4+25 = 5+6+7+8+9$, does that mean that $1=5$ and $2=6$ and ...?
Calling the two sides of your second equation $L_k$ and $R_k$, you are given that $\sum_{k=0}^{n-a} \binom nm L_k = \sum_{k=0}^{n-a} \binom nm R_k$, and you're trying to deduce that $L_k = R_k$. This is just false. The $R_k$ might be the $L_k$ in reverse order, or every $R_k$ might be $0$ except $R_0$ equals exactly what it needs to equal for the first equation to hold, etc.