I want to express the fact that for all $x \in A$ that have the property that for all $y\in x$ $T(x,y)$ is true and there exists an $u \in B$ such that $P(y,u)$ is true AND for all $v\in C$, $Q(y,v)$ and $R(x,v)$ are also true, then $S(x)$ is true.
(Note that this statement is just a toy statement I have invented, because I was not sure I understood this transformation well, so I wanted to make it difficult, keeping with the idea, that if I got this difficult one right, that I probably understood how to do the transformations)
Formulated in a formal language of predicate logic, would this be
$$ \forall x ( x\in A \land \ \ \forall y (y\in x \rightarrow \ ( T(x,y) \land \ldots $$ $$ \ldots \land ( \exists u ( u \in B \land P(y,u) \land \forall v ( v\in C \rightarrow ( Q(y,v) \land R(x,v)))))))\rightarrow S(x) ) $$
(hope I didn't forget any paranthesis...) ?
Is there also a way to move the quantifiers "$\forall$" at the front, so that that string starts with $\forall x \forall y \ldots$ ?
As André said, you can move all of the quantifiers to the front; the result is an expression in what is called prenex normal form, and the conversion can be done quite mechanically. A formula can have more than one prenex normal form, but they will of course all be logically equivalent.
Edit: As Carl reminded me, logically equivalent prenex forms can (contrary to what I originally wrote) begin with different quantifiers: $\forall x \exists y(\phi(x) \land \psi(y))$ and $\exists y \forall x (\phi(x) \land \psi(y))$ are both prenex forms of $\forall x \phi(x) \land \exists x \psi(x)$. However, the order of the quantifiers can also matter: $\forall x \exists y \phi(x,y)$ and $\exists y \forall x \phi(y,x)$ are not logically equivalent. And it is certainly not possible in general to bring all of the universal quantifiers to the front.