Definition of uniform structure From Wikipedia:

A uniform space $(X, Φ) $is a set $X$ equipped with a nonempty family
  $Φ$ of subsets of the Cartesian product $X × X$ ($Φ$ is called the
  uniform structure or uniformity of $X$ and its elements entourages)
  that satisfies the following axioms:
  
  
*
  
*if $U$ is in $Φ$, then $U$ contains the diagonal $Δ = \{ (x, x) : x
    ∈ X \}$.
  
*if $U$ is in $Φ$ and $V$ is a subset of $X × X$ which contains $U$,
  then$V$ is in $Φ$
  
*if $U$ and $V$ are in $Φ$, then $U ∩ V$ is in $Φ$
  
*if $U$ is in $Φ$, then there exists $V$ in $Φ$ such that, whenever
  $(x, y)$ and $(y, z)$ are in $V$, then $(x, z)$ is in $U$.
  
*if $U$ is in $Φ$, then $U^{-1} = \{ (y, x) : (x, y) \in U \}$ is also
  in $Φ$.
  

I was wondering
A. Are 2 and 3 equivalent to that $Φ$ is a filter on $X \times X$?  
B. Is $Δ = \{ (x, x) : x ∈ X \}$ also in $Φ$?
C. How shall I understand 4? 
It reminds me of the triangle inequality in the definition of a
    metric. Are they really related?
Is 4 equivalent to "For any $V$ is in $Φ$, and for any $(x, y)$ and $(y, z)$ in $V$, there exists $U$ in $Φ$ such that $(x, z)$ is in $U$?"
D. in 5, What does it mean that it doesn't require $U \equiv U^{-1}$? To distinguish  distances with different "directions"/"orientations"?
Thanks and regards!
 A: Yes, 2) and 3) just say it is a filter of $X \times X$, all of whose elements contain the diagonal $\Delta$ (axiom 1). This is also often said in the definition (but here they don't want to depend on readers already knowing what a filter is).
The diagonal itself can be in $\Phi$, but then any subset that contains the diagonal will be too (by 2) and this is called the discrete uniformity. This is just $$\Phi = \{ U \subset X \times X \mid \Delta \subset U \},$$ and it will turn out that it induces on $X$ the discrete topology.  
4 is indeed inspired by the triangle inequality, and 5 by the symmetry axiom for a metric ($d(x,y) = d(y,x)$). This is clear if you look at the uniformity induced by a metric $d$, then the uniformity is generated (as a filter base) by the sets $$U_r = \{ (x,y) \in X \times X \mid d(x,y) < r \}$$
If take $U = U_r$ for 4), then we take $V = U_{\frac{r}{2}}$, and note that if $(x,y), (y,z) \in V$, then $d(x,z) <= d(x,y) + d(y, z) < \frac{r}{2} + \frac{r}{2} = r$ so that $(x,z) \in U_r$, as required. 
It's not equivalent to your formulation. For any $V$ that contains the diagonal we can take $$V \circ V = \{ (x,z) \in X \times X \mid \exists y : (x,y) \in V \land (y,z) \in V \}$$ and then $V \subset V \circ V$, so the latter is always in $\Phi$ when $V$ is. So your condition (for any 2 pairs $(x,y), (y,z)$ from $V$, the "composition pair" is in some $U$) is always satisfied from 1 + 2 alone, so adds nothing new.
Note that using the discrete metric $d(x,y) = 1$ when $x \neq y$, $U_1 = \Delta$, making the connection with the discrete uniformity again.
Another example worth looking into is the case of topological groups (a group with a topology that makes the group operations multiplication and inversion continuous), where there are 2 natural uniformities: for any neighbourhood $U$ of the identity $e$ we define one uniformity as generated by all subsets of the form $$\{ (x,y) \in G \times G \mid x\ast y^{-1} \in U\}$$ and the other by all sets of the form $$(x,y) \in G \times G \mid y \ast x^{-1} \in U \}.$$ The axiom 4 then comes down to the fact that for any neighbourhood $U$ of $e$ there is a neighbourhood $V$ of $e$ such that (e.g.) $V \ast V^{-1} \subset U$. which follows from the continuity of the group operations, if you think about it. 
So it's similar to the triangle equality in spirit, and for uniformities that are induced by metrics, it is proved from it, but it's not always of that origin. It's a very important property of uniformities, and allows us to construct continuous functions for the topology induced by a uniformity, similarly to the proof of the Urysohn lemma for normal spaces. It makes such topological spaces completely regular.     
