An "explicit" description of cocomma-categories ? Let the cocomma-square $\require{AMScd}$
\begin{CD}
    A @>T>> Y\\
    @V S V  V \Uparrow_\beta  @VV i_1 V\\
    X @>>i_0> (S\star T)
\end{CD}
where : 


*

*$Ob\,(S\star T) :=Ob\,X\sqcup Ob\,Y$


and $\hom_{(S\star T)}(r,r') $ is given by : 


*

*$\hom_{(S\star T)}(r,r'):=\hom_{X}(x,x')$ if $r=x, r'=x' \in X$,

*$\hom_{(S\star T)}(r,r'):=\hom_{Y}(y,y')$ if $r=y, r'=y' \in Y$,

*$\emptyset$ if $r\in Y$ and $r'\in X$,

*for all $a\in A$, a formal arrow : $Sa \overset{\underline{a}}{\longrightarrow} Ta  $ such that : $id_{Ta}\circ \underline{a}=\underline{a}\circ id_{Sa}=\underline{a}$


all this arrows subject to the relations : 


*

*if $r=x\in X$ and $r'=y\in Y$,  two parallel arrows of $\hom_{(S\star T)}(x,y)$ : 


$$x\overset{f}{\longrightarrow}Sa\overset{\underline{a}}{\longrightarrow}Ta\overset{g}{\longrightarrow}y$$
$$x\overset{f'}{\longrightarrow}Sa'\overset{\underline{a}'}{\longrightarrow}Ta'\overset{g'}{\longrightarrow}y$$
are equal if exists a "zig-zag" in $A$ : 
$$a\overset{q_{1}}{\longleftarrow}a_{1}\longrightarrow...\overset{q_{n}}{\longrightarrow}a' $$
and arrows $f_{i}:x\longrightarrow Sa_{i}$ in $X$ and $g_{i}:Ta_{i}\longrightarrow y$ in $Y$ making the diagram commutes  :
\begin{CD}
x @>f>>    Sa @>\underline{a}>> Ta @>g>> y \\
 @|   @A Sq_1 AA         @AA Tq_1 A  @| \\
x @>f_1>>  Sa_1 @>>\underline{a_1}> Ta_1   @>g_1>> y \\
@|   @V Sq_2 VV   @VV Tq_2 V @| \\
\cdots @>>> \cdots @>>> \cdots   @>>> \cdots \\
@| @V Sq_n VV   @VV Tq_n V @| \\
x @>>f'>  Sa' @>>\underline{a'}> Ta' @>>g'> y\\
\end{CD}
In this description : $\beta_a:=\underline{a}$.
Can someone confirm me if it is right or not ?  
 A: Your construction is basically correct.  The only slight mistake (probably you had the correct thing in mind and this was just a mistake in phrasing) is that the formal arrows $\underline{a}$ are not sufficient to give the hom-sets $\hom_{S\star T}(r,r')$ when $r\in X$ and $r'\in Y$; you also need the "formal composites" of these arrows with arrows in $X$ and $Y$ on either side, which then get quotiented by the equivalence relation you describe.
Here is an abstract argument for why this construction is correct.


*

*Every comma object is a discrete two-sided fibration.

*Dually, every cocomma object is a codiscrete two-sided cofibration.

*In Cat, every discrete two-sided fibration is the comma of its cocomma, and every codiscrete two-sided cofibration is the cocomma of its comma.  This sets up an equivalence between discrete two-sided fibrations and codiscrete two-sided cofibrations, both of which represent profunctors from $Y$ to $X$.

*Moreover, the comma of the cocomma is a reflection from arbitrary spans into discrete two-sided fibrations, and dually the cocomma of the comma is a coreflection from arbitrary cospans into codiscrete two-sided cofibrations.  Thus, to compute an arbitrary cocomma, it suffices to compute the reflection of a span into discrete two-sided fibrations, regard it as a profunctor, and then take the collage of that profunctor.

*Every span $X \leftarrow A \to Y$ generates a free two-sided fibration $X \leftarrow X^{\mathbf{2}}\times_X A \times_Y Y^{\mathbf{2}} \to Y$, where $X^{\mathbf{2}}$ is the category of arrows in $X$.  The objects of this free two-sided fibration are therefore triples $(a,f,g)$ where $a\in A$, $f:x\to S a$, and $g:T a \to y$, which we can regard as the above "formal composites".  Now it suffices to take the discrete reflection of this over $X$ and $Y$, which amounts to imposing your zig-zag equivalence relation.

A: There is an explicit description of colimits in $\bf Cat$: this, however, does not address your question since you are coping with a lax colimit.
Since I am a rather stubborn when it comes to using coend calculus to clarify something, I'm here again with a suggestion on how to look at your construction in a way I like.
Of course, Mike's answer is perfectly fine: this is an equivalent, if not more verbose, way to present your $S\star T$. 
Since this construction, as a lax (or colax? Mike had a wonderful mnemonic trick to distinguish the two) pushout, is not symmetric in $S,T$, and since this reflects a morphological principle I really care about[1], I'll stick to denote your $\star$ as $\ltimes $
==
Right before the section "Using comma object" on the $n$Lab page about exact squares there is the claim

the cocomma object of a cospan $Y\overset{u}{\leftarrow} A \overset{f}{\to} X$ is precisely the cograph of the profunctor $Y(u,1) \circ X(1,f)$. 

Let's prove this statement. First of all we have to look closer at the definition of the "cograph": following again $n$Lab this is the category having objects the disjoint union $B\coprod C$ and arrows those of $B$, in the copy of $B$, those of $C$, in the copy of $C$, no arrows between $c$ and $b$, and exactly as many arrows as 
$$
\int^a Y(ga,c)\times X(b, fa)
$$
between $b\in B$ and $c\in C$. This coend powerfully encodes all the cumbersome generators of equivalence relations you would have to write if you didn't know how to write an integral symbol :-)
The intuition for what's into $\hom_{X\ltimes Y}(b,c)$ is precisely the same as above! It's a set of "fake arrows"
$$
b \xrightarrow{\varphi} fa \overset{(\xi)}\rightsquigarrow ga \xrightarrow{\psi}c
$$
where the arrow $(\xi)$ exists "morally". I suspect that (and there's some homotopy theory going on here, I think[2]) you would have obtained the same category up-to-equivalence defining $X\ltimes Y$ to be the strict pushout
$$
\begin{array}{ccc}
A \coprod A & \to & A \times {\bf 2}\\
\downarrow && \downarrow \\
X\coprod Y & \to & P
\end{array}
$$
if you look closer you'll see what is really "creating" the fake arrows, but my point is that you shouldn't because coends are far better behaved quotients. :-)
Inside the coend, you identify two arrows $b \xrightarrow{\varphi} fa \overset{(\xi)}\rightsquigarrow ga \xrightarrow{\psi}c$ and $b \xrightarrow{\varphi} fa' \overset{(\xi')}\rightsquigarrow ga' \xrightarrow{\psi}c$ precisely when there is a zig-zag $a\leftrightarrows a'$ fitting in the diagram with which you opened the thread.
==
Now we have enough material to fiddle and find the universal property: a natural transformation $\zeta\colon i_0 S \Rightarrow i_1 T$ comes from the tautological equivalence class of identities in the following way.
Since $\zeta_a : i_o fa \to i_1 ga$ is an arrow $fa \to ga$ when the two objects are regarded in $X\ltimes Y$, we have to find a distinguished element in $\hom_{X\ltimes Y}(fa, ga)$, or rather in the coend above where $c=ga, b=fa$. This gives the image of $(1_{ga}, 1_{fa}) \in Y(ga,ga)\times X(fa,fa)$ under the composition
$$
\begin{array}{ccc}
\hom(ga,ga)\times \hom(fa,fa) &\ni&(1_{ga}, 1_{fa})\\
\downarrow\\
\coprod_{a\in A}\hom(c,ga)\times \hom(fa, b)\\
\downarrow\\
\hom_{X\ltimes Y}(fa, ga) &\ni& [(1_{ga}, 1_{fa})]
\end{array}
$$
as a natural candidate for $\zeta_a \colon fa\to ga$. And in fact this works! 
Suppose you are given a commutative square
$$
\begin{array}{ccc}
A &\to & B \\
\downarrow &\Downarrow  & \downarrow\\
C &\to&Y
\end{array}
$$
filled by a 2-cell $\theta : wf \Rightarrow vg$. Then define a unique functor $u : X\ltimes Y \to Y$ on objects and true arrows in $B$ or $C$ acting as $v\colon C\to Y, w\colon B\to Y$; the action of $u$ on a fake arrow
$$
b \xrightarrow{\varphi} fa \overset{(\xi)}\rightsquigarrow ga \xrightarrow{\psi}c
$$
is induced by the composition
$$
wb \xrightarrow{w\varphi} wfa \xrightarrow{\theta_a} vga \xrightarrow{v\psi}vc
$$
(all arrows exist now!), and I'm more than happy to leave you a bunch of routine checks ($u$ is unique due to the tautological definition of $\zeta$; $\zeta$ has a 2-dimensional universal property as well; $\zeta\circ -$ reflects isos).
Some obvious remarks:


*

*Flip $X$ and $Y$, you obtain the opposite construction with a 2-cell in the opposite direction.

*That's awesome! The cograph of a profunctor is the category of elements of the same profunctor regarded as a presheaf $B^\text{op}\times C \to {\rm Set}$. This reinforces the inuition of $\text{Elts}(F)$ as a suitable lax colimit of the presheaf $F$. The consequences of this never stop to fill me with wonder.

*What if $f$ or $g$ is the identity functor? You get the category of elements of $Y(g,1)$ or $X(1,f)$, which is...

*What's the cograph of $\hom(f,g)$ and how does it fit into this constructions?


===
[1] i.e. the idea that (non-)commutative operations deserve (non-)symmetric symbols.
[2]there's some directed homotopy theory going on, to be precise. See Grandis, "Directed Algebraic Topology: Models of Non-Reversible Worlds"
Volume 13 di New Mathematical Monographs.
