I haven't looked up references, but for 2-dimensional linear groups, the name of L.E. Dickson comes to mind. For three dimensional groups, the names of Mitchell,Bloom and Hartley come to mind. In the latter case, I think there is a Trans AMS paper in the late (19)60s which does the 3-dimensional groups, and may have references for the 2-dimensional case.
To some extent, this may depend on the level of detail you want. If you are prepared to ignore factor groups of order prime to $p$, you might as well look at finite subgroups of ${\rm PSL}(2,K)$ generated by unipotent elements. If $p$ is odd, these groups have dihedral Sylow $2$-subgroups. If $p=2$, they have elementary Abelian Sylow $2$-subgroups. Therefore, any of their finite subgroups would be considered to be "known" by finite group theorists, but maybe not in the detail you want, so I'll go a bit further. Let us consider irreducible subgroups (otherwise we are looking at triangularizable groups which have a normal Sylow $p$-subgroup). For the moment, let's suppose that $p >3$. Then if $G$ is a finite irreducible subgroup of ${\rm SL}(2,K)$ generated by its unipotent elements, the Hall-Higman theorem tells us that all normal subgroups of order prime to $p$ are central. Any normal subgroup of order a power of $p$ is trivial by irreducibility. Then by general group theory, $G$ has a quasi-simple normal subgroup $L$ (that is,$L = [L,L]$ and $L/Z(L)$ is simple). Furthermore $G/L$ is isomorphic to a subgroup of ${\rm Out}(L)$. Now (as $p>3$), $L/Z(L)$ has dihedral Sylow $2$-subgroups, so is listed (in papers of Gorenstein-Walter, and Bender-Glauberman, well before the full classification of finite simple groups). The possibilities are usually of the form ${\rm PSL}(2,r)$ for some power of a prime $r$ (exceptions such as $A_{7}$ and one of the Janko groups don't arise here because of the low-dimensional representation). If $r$ is a power of a prime $q \neq p$, then $L$ has cyclic Sylow $p$-subgroups of order $p$ of index $2$ in their normalizers. But also $L/Z(L)$ has a subgroup isomorphic to ${\rm PSL}(2,p)$ (by an argument which really dates back to Dickson). Thus $p \leq 5$, and I think these cases can be dealt with (although we have assumed $p >3$, it is worth pointing out that ${\rm PSL}(2,5)$ does occur as a subgroup of ${\rm PSL}(2,9)$), so that (with few, if any) exceptions, $L/Z(L)$ is isomorphic tp ${\rm PSL}(2,t)$ for $t$ some power of $t$.
If $p = 2$, then every finite subgroup $G$ of ${\rm PSL}(2,K)$ has a $TI$ Sylow $2$-subgroup. That is, we have $S \cap S^g =1$ for each Sylow $2$-subgroup $S$ and each $g$ outside $N_G(S).$ If $G$ does not have a cyclic Sylow $2$-subgroup, then by Theorems of Bender and Suzuki, there is a normal subgroup $N$ of odd index in $G$ such that $N \equiv {\rm SL}(2,2^n)$.
The case $p=3$ can be dealt with in a similar fashion, but there are extra exceptions, such as ${\rm PSL}(2,5)$, and problems caused by the non-simplicity of ${\rm PSL}(2,3).$