As I was trying to understand the category $Ind(C)$ of diagrams of the form $I \to C$, where $I$ is a small filtered $(0,1)$category, I wondered whether it is possible to define morphisms directly, without the limitcolimit construction (see https://ncatlab.org/nlab/show/indobject), and then try to show that its equivalent to the common definition as $\text{lim$_i$ colim$_j$}\,C(F(i),G(j))$. A natural choice would be to define a morphism $(F:I\to C) \to (G:J \to C)$ as a pair $(f,g)$, where $f:I \to J$ is a functor in the $(1,2)$category of all posetcategories, and $g:F \Rightarrow G \circ f$ is a natural transformation. Then, the functor $C \hookrightarrow Ind(C)$ is clearly fully faithful. While trying to show that each object in $Ind(C)$ is a colimit of itself, I recognized that I should rather look at 2colimits, but then I should define what a 2morphism is. I tried several things but I'm not an expert for higher categories, so unfortunately I lack the necessary intuition... do you know a canonical choice for 2morphisms in $Ind(C)$?

2$\begingroup$ This paper might be relevant to your question: arxiv.org/abs/1406.6229 $\endgroup$ – Yonatan Harpaz Apr 27 '16 at 20:44

$\begingroup$ Thx! This is exactly the kind of result I desired! $\endgroup$ – Bipolar Minds Apr 27 '16 at 21:38
On a more general note, you're (gradually) building a subcategory of the 2fiber $\textbf{Cat}/C$. An excellent reference for those is the following paper, which helped me a lot when I was looking into this stuff:
A Bullejos and M Cegarra, On the geometry of 2categories and their classifying spaces, KTheory 29 (2003) 211229.
The journal where this paper appeared is now famously defunct, but you can find a pdf of the article here. The explicit answer to your question regarding 2morphisms is as follows. Let $F$ and $G$ be as in your question and consider a pair of 1morphisms $(f,g)$ and $(f',g')$, also as in your question. A 2morphism $(f,g) \Rightarrow (f',g')$ is a natural transformation $\beta:f \Rightarrow f'$ (i.e., a 1morphism in the category of functors from $I$ to $J$) so that the following holds: $$(G\circ\beta) * g = g'$$
Here $*$ stands for vertical composition while $\circ$ is horizontal composition. Pictorially, we have:

$\begingroup$ Wow, thx a lot, I will check that out! $\endgroup$ – Bipolar Minds Apr 28 '16 at 17:41