Okay so it's not actually hard to get either $f(X)$ or $g(X)$, you just need to keep on using the well known generating function which I mentioned earlier.
Above, we have already shown that the generating function for $f$ is $$f(X)=\frac{X^{2m+1}}{(1-2X)^{m+1}}$$We will show that the generating function for $g$ is the same. Rewrite $g$ as
$$g(X)=\frac{X^{m+1}}{2^m}\sum_{n\geq 0}\binom{n-m-1}{m}(2X)^{n-m-1}=\frac{X^{m+1}}{2^m}\sum_{j\geq 0}\binom{j}{m}X^j$$
Now we can use that generating function to get$$g(X)=\frac{X^{m+1}}{2^m}\frac{(2X)^m}{(1-2X)^{m+1}}=\frac{X^{2m+1}}{(1-2X)^{m+1}}$$
Equating coefficiants of the generating functions tells us the left hand side and right hand side of the original equaion are indeed equal for all $n$.
