Warning: the following computations seem to be confirmed by Maple, though they are now a bit fishy to me. I leave them for reference.
Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:
\begin{equation} \sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k} \end{equation} and \begin{equation} \sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}\\ . \end{equation}
According to Maple, these sums to:
\begin{equation} (-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1} \end{equation} and \begin{equation} (-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2} \end{equation} The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series that you gave:
\begin{equation} \sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)} \end{equation} and \begin{equation} \sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\\ . \end{equation} As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$. In case, I will edit later and add the details.