There is indeed a pretty simple way to derive necessary and sufficient conditions on $(g,n_1,...,n_k)$ for existence of such a hyperbolic structure on your surface $S$.
To derive a necessary condition, let $S$ be your surface equipped with such a hyperbolic structure. Let $DS$ be obtained by doubling $S$ across the boundary, in other words $DS$ is the quotient of a pair of copies of $S$ by identifying the two copies of the boundary using the identity map. The surface $DS$ is a surface of some finite genus $g'$ and some finite number of punctures $p'$, and one can work out formulas for $g'$ and $p'$ in terms of $(g,n_1,...,n_k)$ (I'll indicate more details below of how one does this). Furthermore, by doubling the hyperbolic structure on $S$ one sees that $DS$ possesses a hyperbolic structure, i.e. a complete hyperbolic metric.
Just as for compact surfaces, existence of a complete hyperbolic metric on a punctured surface such as $DS$ is equivalent to saying that $DS$ has negative Euler characteristic:
$$\chi(S) = 2 - 2g' - p' < 0
$$
It turns out that this necessary condition is also sufficient, because for any complete hyperbolic structure on $DS$, under the embedding $S \hookrightarrow DS$ one can straighten the boundary of $S$ to get the desired hyperbolic structure on $S$.
The number of punctures on $DS$ is simply
$$p' = \sum_{i=1}^k n_i
$$
because there is exactly one puncture on $DS$ for each boundary point that was removed to form $S$.
To compute genus, its easier to work with Euler characteristic, and we have
$$\chi(S) = 2 - 2g - k - \sum_{i=1}^k n_i
$$
The disjoint union of two copies of $S$ has twice the Euler characteristic:
$$\chi(S \cup S') = 4 - 4g - 2k - 2 \sum_{i=1}^k n_i
$$
Identifying two circle components of the boundaries does not change Euler characteristic, but identifying two arc components reduces the Euler characteristic by $1$:
$$\chi(DS) = 4 - 4g - 2k - 3 \sum_{i=1}^k n_i + \#\{i \mid n_i=0\}
$$
Since $\chi(DS) = 2 - 2g' - p'$, and since we have a formula for $p'$, we can solve for $g'$.
