Can I do instantiation like this? Suppose, if I have been given this: $\forall x \in A(P(x))$ and $\exists y \in A(Q(y))$.
Now from $\forall x \in A(P(x))$ using universal instantiation, I get $P(c)$ where $c$ is an arbitrary element in $A$. Now my doubt is, if using $\exists y \in A(Q(y))$, can I conclude $Q(c)$ ?. My reasoning says yes, because $c$ is an arbitrary element in $A$ and $y \in A$. Is this type of instantiation from universal to existential has a special name ?
 A: No, you have to go the other way around: from $\exists y\in A(Q(y))$ we instantiate to get a $c$ with $Q(c)$; we now conclude that $P(c)$ holds, because $\forall x\in A(P(c))$.

One way to picture this is as a game. You're trying to show that a statement - say, $\exists x\in A(P(x) \wedge  Q(x))$ - is true, and your opponent (me) is trying to thwart you. Your example goes as follows:


*

*You: "I know that $\forall x\in A(P(x))$. Give me some $x\in A$ such that $P(x)$."

*Me: "Sure, have this thing $c$. I promise you $P(c)$ holds."

*You: "I know that there is some $x$ with $Q(x)$. I claim $Q(c)$ holds, in particular, so I've found an $x$ with $P(x)\wedge Q(x)$."

*Me: "Haha, I chose a $c$ that doesn't satisfy $Q(c)$! You have made an incorrect deduction, so you lose."
The right way to approach this is the opposite:


*

*You: "I know $\exists x\in A(Q(x))$. Give me some $x\in A$ with $Q(x)$."

*Me: "Sure, have this thing $c$. I promise you $Q(c)$ holds."

*You: "Well, I know that $\forall x\in A(P(x))$ - so in particular, $P(c)$. So I have $P(c)\wedge Q(c)$."

*Me: "Curses!"
The general name for this sort of thing is game semantics. I find it a very helpful framework for understanding the (in)validity of logical arguments (and many other things besides).
A: 
Now from $∀x∈A(P(x))$  using universal instantiation, I get $P(c)$  where $c$  is an arbitrary element in A . Now my doubt is, if using $∃y∈A(Q(y))$ , can I conclude $Q(c)$  ?.

No.

My reasoning says yes, because $c$  is an arbitrary element in $A$  and $y∈A$ . Is this type of instantiation from universal to existential has a special name ?

Your reasoning is almost correct.   You can pick a value $c$ where $Q(c)$ holds, but this not an arbitrary value from a universal instantiation of the first statement.
It's a witness from an existential instantiation of the second statement. 
You can then select this witness from a universal instantiation, but, as it's not arbitrary you cannot subsequently use it in universal generalisation.   A witness can only be used in a subsequent existential generalisation.
Thus we do so:
$\begin{align}
\exists y\in A\;Q(y) \;&\vdash\; Q(c)&\text{EI}, c\in A 
\\[2ex]
 \forall x\in A\;P(x) \;&\vdash\; P(c)&\text{UI}(\mid_c^x)
\\[1ex]
 P(c), Q(c) \;&\vdash\; P(c)\wedge Q(c)&\wedge\text{I}
\\[1ex]
 P(c)\wedge Q(c) \;&\vdash\; \exists x\in A\;(P(x)\wedge Q(x))&\text{EG}
\\[2ex]
\therefore \forall x\in A\;P(x),\, \exists y\in A\;Q(y)\;&\vdash\; \exists x\in A\;(P(x)\wedge Q(x))
&
\Box
\end{align}$
