$1)$ The connection should be clear after reading the Wikipedia article carefully, especially the examples. The index category encodes the shape of the diagram, whereas the functor itsself says which morphisms are attached to which edges of the shape. And yes, every shape is realized by some category. In fact, more generally, every graph can be seen as a category, where the vertices become the objects and the paths become the morphisms.
For example, the graph
$$\require{AMScd}
\begin{CD}
a@>>> b\\
@VVV @VVV\\
c@>>> d
\end{CD}$$
gives a index or shape category with four objects $a,b,c,d$ and six non-identity morphisms $a \to b, b \to d, a \to c, c \to d, a \to b \to d$ and $a \to c \to d$. This also describes the diagrams on this shape category. It is commutative iff $a \to b \to d$ and $a \to c \to d$ become evaluated to the same morphism.
$2)$ The assertion "a diagram is commutative iff the source is a poset?" is clearly wrong. But "if" is correct, for trivial reasons. In a poset, all parallel morphisms are equal, so the commutativity condition is quite empty.
$3)$ Yes. A category is a poset iff every diagram in that category commutes. When one starts to learn the basic notions of category theory, it is often helpful to look at posets first, because then many things simplify, but the main ideas remain. For example, colimits become suprema, limits become infima, functors become monotonic maps, adjunctions become Galois connections, etc.