$f:X\rightarrow Y$ is an epimorphism iff first diagram on the picture is a pushout.
Can I say "there is a pushout along the morphism h"?
Is my example of non-epimorphism (below on the picture) correct?
(I want to see example of non-uniqueness of morphism $h$, but my example contradicts with wikipedia's "all directed paths in the diagram with the same start and endpoints lead to the same result by composition" in the bottom triangle )
(I'm not sure, but) How can we correct this example to make it "fully categorical"(not dependent on sets: {a},{1,2})? Can we?
p.s. LaTeX code is in the comment inside question
The short answer to your question: "Is my example of non-epimorphism (below on the picture) correct?" is no (explanation below).
The short answer to your question: "Can I say "there is a pushout along the morphism h"?" is no again (sorry :-)). This terminology is used for pushbacks AFAIK and is always referring to one of the sides of the square, not to the "mediating" morphism $h$.
Your bottom diagram is ok by itself, provided you understand and warn your readers that it is not a commutative diagram. Non-commutative diagrams are less useful and perhaps even a bit misleading for the inexperienced user (since we are all more used to deal with commutative diagrams), still they can be a visual help to see where all the various morphisms go.
I shall use categorical language, to answer your...nth question :-)
The top diagram is perfect. Let's examine the bottom non-commutative diagram. Your goal is to show with a counterexample that a given cocone (the pushout square :{a},{1,2},{1,2},{1,2}) is not a colimit. You can achieve this goal by showing that there is a cocone that is either not connected or connected by more than one morphism in the cocone category (the contrary of "exactly one morphism" is "zero or more-than-one morphism"). Please note I said "morphism in the cocone category", not just any morphism connecting the cocone vertices. These morphisms respect the cone structure In your bottom diagram you have 2 cocones: the "square" and the "quadrilateral". They are exactly the same cocone, so obviously they can be connected by an ìdentity, which respects the cone structure. Your mistake is to consider the other morphism: $1 \mapsto 1$, $2 \mapsto 1$. This is not a "morphism in the cocone category", precisely because- when composed with the sides of one cocone (the square ids)- it does not commute with the sides of the other cocone (the quadrilateral ids). As a matter of fact, you chose a bad counterexample because there are no other morphisms in the cocone category, connecting the given cocone (the square) with itself.
How to fix it? Just take another cocone (quadrilateral). If you observe your top diagram, you see that $q_1 = q_2=h$, because of the ids in the square. So if you want a counterexample, just find different qs. For example: replace the lowest id ($q_2$ in the top diagram) with your $1 \mapsto 1$, $2 \mapsto 1$ and leave $q_1=id$ . Now there is no possible morphism $h$ in the cocone category connecting the square and the quadrilateral. So the square is not a colimit.
You can read about the category of cones and morphisms of cones in Awodey , for example