Specifically if I have a known function $F(s)$ is there some way I can find a function $f(n)$ that satisfies this equation?
$$F(s) = \sum_{n=1}^\infty \frac{f(n)}{n^s}$$
I'm imagining something similar to finding the coefficients of a Fourier series.
On
You can do this using Perron's formula. It determines all but a finite number of $f(n)$, the rest of which we can find using a Vandermonde-argument:
If $F$ has finite abscissa of (absolute) convergence $\sigma_a$ and $n>\sigma_a$ is an integer, using Perron's formula we can express $f(n)$ as a difference of contour integrals (with values $\sum_{k<n}f(n)$ and $\sum_{k\leq n}f(n)$). Having determined those $f(n)$, subtract $\sum_{n>\sigma_a}f(n)/n^s$ to reduce to the case where only finitely many $f(n)$ are nonzero. Evaluating the series at $s=1,2,3,\ldots$ gives a Vandermonde linear system in the $f(1),f(2),\ldots$ which determines $f$.
Of course, if $\sigma_a=+\infty$, as a function $F(s)$ carries no information.
Although there might be some trouble in implementing this in practice, I had found a trick that same afternoon that seemed to work. I defined this finite difference operator:
$$\square_{a,b} F(s) = \frac{F(s-1)-bF(s)}{a-b}$$
It "kills" the $b$th term in the Dirichlet series while leaving the $a$th term untouched. In order to extract the term I want, I do this for all values of $b \ne a$ and in the end I get this expression for $f(a)$ given $F(s)$:
$$f(a) = \left(a^s\prod_{b=1, b \ne a}^\infty \square_{a,b} \right)F(s) $$