N songs, M instruments, how to pick K instruments to cover maximal amount of songs
I don't know a canonical name for this problem, but unfortunately it's NP-hard, even when every song requires at most 2 instruments.  The upshot is that it's extremely unlikely that any polynomial-time algorithm exists that will solve every instance of your problem.  I'll show this by reduction from the NP-hard problem Maximum Clique, in which we are given an undirected graph $G = (V, E)$ and a positive integer $k$, and asked whether there exists a clique of size $k$ in the graph.

To construct an instance of your problem from the Maximum Clique Instance: Create a song $s_{uv}$ for every edge $(u, v)$ in $E$, and an instrument $t_v$ for every vertex $v$ in $V$, and make song $s_{uv}$ require the two instruments $t_u$ and $t_v$.  Set $K$ (part of the input to your problem) equal to $k$ (part of the input to the Maximum Clique problem).  Now run any algorithm that solves your problem on the just-created instance: if the answer is $\ge k(k-1)/2$ then return YES, otherwise return NO.  (Clearly this procedure could work just as easily with an algorithm that solves the "pure decision" form of your problem, i.e., when a number $r$ is also given as part of the input, and the task is to determine whether it is possible to choose at most $K$ instruments such that at least $r$ songs can be played.)

It is easy to see that, if $G$ contains a clique of size at least $k$, then we must correctly get a YES answer: by assumption, there is a $k$-vertex clique, which comprises $k(k-1)/2$ vertex pairs, each of which are connected by an edge, so choosing the corresponding $k$ instruments satisfies the $k(k-1)/2$ corresponding songs.

In the other direction, suppose that the algorithm for your problem reports that at least $k(k-1)/2$ songs can be played; we need to show that this implies that there is a $k$-clique in $G$.  Given that we permitted at most $k$ instruments to be chosen, there certainly cannot be more than $k(k-1)/2$ songs that we can play, since every song requires a distinct pair of instruments: so we can infer that exactly $k(k-1)/2$ songs can be played.  But every choice of at most $k$ instruments that does not correspond to a $k$-clique in $G$ omits at least one of these $k(k-1)/2$ edges, and thus allows strictly fewer than this maximum possible number of songs to be played; thus the only subgraphs on at most $k$ vertices that can result in at least $k(k-1)/2$ edges are subgraphs that are $k$-cliques.