About the notions of Grothendieck Universe and Tarski Universe I assume ZFC.
Let $U$ a set with the following (1), (2), (3): 
1) $\omega\in U$
2) $x\in U\ \Rightarrow x\subset U$
3) $x\in U\ \Rightarrow \mathcal{P}(x)\in U$  (where $\mathcal{P}(x):=${$y| y\subset x$})
Then  by  these   premises, I wish prove that $(4)\Leftrightarrow (4')$ where:
4) if $x\subset U$ and $|x|<|U|$ then $x\in U$  (where $|a|$ is the cardinality of the set $a$)
4') If $f: a\to U$ is  function and $a\in U$ then $\bigcup_{s\in a}f(s)\in U$ .
The Book of Monk "Introduction to set theory' claim this equivalence as a exercise.
and  the implication $(4)\Rightarrow (4')$ is immediate from the book above..
I tried to prove by induction that $|U|\subset U$, or tried to generalize the Mostowsky theorem for get a  inijection $|U|\to U$ which preserves the relation '$\in $' (unsuccessfully). I have see also some posts here about this problem, but I hope exist  a (relatively) simple answere.. 
THen I ask: How to prove $(4')\Rightarrow (4)$ ?
 A: I don't see a really slick proof, so here's a rough (and fairly standard) one.  Assume that $U$ satisfies (1,2,3,4').  Notice that, by (2) and (3), $U$ is closed under subsets, so the cardinalities of elements of $U$ form an initial segment of the cardinal numbers; let $\kappa$ be the smallest cardinal not in this initial segment.  It follows from (1) that $\kappa$ is uncountable, from (3) that whenever $\mu<\kappa$ then $2^\mu<\kappa$, and from (4') that $\kappa$ is a regular cardinal.  So $\kappa$ is strongly inaccessible.  If $U$ contained any sets of rank $\geq\kappa$, then the lowest-rank such sets would, by (2), have rank exactly $\kappa$; but this is impossible, because (by regularity of $\kappa$) any set of rank exactly $\kappa$ must have cardinality at least $\kappa$.  The conclusion, therefore, is that all sets in $U$ have rank $<\kappa$.  With the usual notation for the levels of the cumulative hierarchy, we have $U\subseteq V_\kappa$.
Next, notice that for all $\alpha<\kappa$, $V_\alpha$ has cardinality $<\kappa$.  (This is proved by induction on $\alpha$, using at successor stages that $\mu<\kappa$ implies $2^\mu<\kappa$, and using at limit stages that $\kappa$ is regular and uncountable.)  Therefore, $V_\kappa$, being the union of $\kappa$ sets $V_\alpha$ each smaller than $\kappa$, has cardinality only $\kappa$.  In view of the result in the preceding paragraph, we have $|U|\leq \kappa$.
This means that, if $x$ is as in the statement (4) that we want to prove, then $|x|<\kappa$, and therefore there exists some $a\in U$ with $|a|=|x|$.  Fix such an $a$ and fix a bijection $f:a\to x$.  Applying (4'), we find that the union $y$ of all the members of $x$ is an element of $U$.  But each member of $x$ is a subset of $y$ and therefore an element of the power set of $y$.  So we have $x\subseteq\mathcal P(y)$ and, by (3), $\mathcal P(y)\in U$.  We already noted above that $U$ is closed under subsets, so it follows that $x\in U$, as desired.
