\begin{align*}
\sum_{x=0}^n \binom{n}{x} y^x & = (1+y)^n, \text{ by the Binomial Theorem}\\
& = (1+y)^{n_1} \cdot (1+y)^{n_2} \\
& = \sum_{x=0}^{n_1} \binom{n_1}{x} y^x \cdot \sum_{x=0}^{n_2} \binom{n_2}{x} y^x \\
& = \sum_{x=0}^n \sum_{t=0}^r \binom{n_1}{r} \binom{n_2}{r-t} y^x
\end{align*}
Equating like coefficients for the various powers of y gives
\begin{equation*}
\binom{n}{r} = \sum_{t=0}^r \binom{n_1}{r} \binom{n_2}{r-t}.
\end{equation*}
Dividing gives
\begin{equation*}
1 = \sum_{x=0}^r f(x).
\end{equation*}