This should be a comment - but it is too long:
Assume that $U = U_1 = U_2$. I want to show that $\mathbb{P}_U ^2 \cong \mathbb{P}_U\times \mathbb{C}$ where $\mathbb{C}$ is the Cohen forcing.
Let $\{ {\alpha^0}_i\}_{i <\omega}, \{ {\alpha^1}_i \}_{i<\omega}$ be the two Prikry sequences.
Set $\{ \gamma_n \}_{n<\omega} = \{ {\alpha^0}_i\}_{i <\omega} \cup \{ {\alpha^1}_i \}_{i<\omega}$ (the Prikry sequence) and $f:\omega \rightarrow P(2)\setminus \{\emptyset\}$, $f(n) = \{ i < 2 | \gamma_n \in \{ {\alpha^i}_m \}_{m < \omega} \}$ (the Cohen real).
This gives us an isomorphism: send conditions from the dense set $( \langle s_0, A\rangle , \langle s_1, A\rangle )$ (the same $A$, $\min A > \max s_0 \cup s_1$) in $\mathbb{P}_U^2$ to $(\langle s_0\cup s_1, A \rangle, f\restriction |s_0 \cup s_1|) \in \mathbb{P}_U \times \mathbb{C}$ (we can calculate $f\restriction |s_0 \cup s_1|$ since we can only add elements in the Prikry sequence above the $\max s_0 \cup s_1$). This is an order preserving bijection between those two posets.
Edit: The answer for Question 1.2 depends on the exact support that you're using:
For finite support - $\mathbb{P}^\omega$ trivially collapses $\kappa$ to $\omega$ - the function that assign to each $n$ the first element in the $n$-th Prikry sequence is onto $\kappa$.
For full support - $\mathbb{P}^\omega$ also collapses $\kappa$: in the generic extension there is a surjection from $(2^{\aleph_0})^V$ to $\kappa$.
Choose a $\omega$-Jonsson function on $\kappa$, i.e. function $f:[\kappa]^\omega \rightarrow \kappa$, such that for every $x\subset \kappa, |x|=\kappa$, $f^{\prime\prime}([x]^\omega) = \kappa$. Let $\{\alpha^j_i \}_{i<\omega}$ be the Prikry sequence added by the $j$ component of $\mathbb{P}^\omega$.
We define a function $g: (\omega^\omega)^V \rightarrow \kappa$ by $g(z) = f(\{\alpha_{z(n)}^n | n < \omega\})$ .
I claim that $g$ is surjective: Let $p = \langle g_i, A_i | i < \omega \rangle \in \mathbb{P}^\omega$ and $\alpha < \kappa$. WLOG, $A=A_i$ for every $i$. Choose $y\in A^\omega$ such that $f(y) = \alpha$ and extends each $g_i$ by the corresponding element of $y$. Since the sequence of lengths of $g_i$ is real from $V$ - the new condition forces $\alpha \in \text{im }g$.
Since we're dealing with Prikry forcing there is at least one more support that we should consider - the Magidor support, namely the conditions are all elements of the form $\langle g_i, A_i |i < \delta\rangle$ such that $\{i<\delta | g_i \neq \emptyset \}$ is finite. In this case (as long as $\delta < \kappa$), we can apply the same idea as above and get that $\mathbb{P}^\delta \cong \mathbb{P}\times \mathbb{D}$ where $\mathbb{D}$ is a forcing that adds a generic function from $\omega$ to the set of all finite, non empty subsets of $\delta$, so as long as $\delta < \kappa$ - $\kappa$ is not collapsed.
When $\delta = \kappa$ this argument doesn't work, so I don't know if $\kappa$ is collapsed by the Magidor power $\mathbb{P}^\kappa$ or not.
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Remark 1. The answer to question 1.1 is yes, even if $U_1\neq U_2.$
Theorem. Let $U,V$ be normal measures on $\kappa.$ Then forcing with $\mathbb{P}_U\times \mathbb{P}_V$ preserves all cardinals.
Proof. It suffices to consider the case where $U$ is not equal to $V$. So let $A^*\in U$ such that $\kappa-A^*\in V.$ Let $W=\{ X\subset\kappa: X\cap A^*\in U, X\cap (\kappa-A^*)\in V \}.$ It is easily seen that $W$ is $\kappa-$complete filter on $\kappa$ which is Rowbottom (for any $f: [D]^{<\omega}\to \lambda<\kappa, D\in W$, there is $E\in W, E\subset D$ such that $card(f'' [E]^{<\omega}) \leq \omega$ ). So we can define $\mathbb{P}_W$ and by Devlin's paper "Some Remarks on Changing Cofinalities"1, forcing with $\mathbb{P}_W$ preserves cardinals. As above argument, we have a forcing isomorphism from the dense subset $\{((s, A)(t, B))\in \mathbb{P}_U\times \mathbb{P}_V: A\subset A^*, B\subset \kappa-A^*, max(s\cup t)< min(A), min(B) \}$ of $\mathbb{P}_U\times \mathbb{P}_V$ to $\mathbb{P}_W\times \mathbb{C}$ given by $((s, A)(t, B))\to ((s\cup t, A\cup B), f\restriction |s\cup t|),$ where $f$ is defined as above argument. The result follows.
- Keith J. Devlin, Some remarks on changing cofinalities, J. Symbolic Logic 39 (1974), 27--30.
Remark 2. $\kappa$ is collapsed by the Magidor power $\mathbb{P}^\kappa$, by the following argument:
For any $\delta<\kappa,$ we may factor $\mathbb{P}^\kappa$ as $\mathbb{P}^\delta \times \mathbb{P}^{\kappa -\delta}$, so by the above argumets we can conclude that all cardinals $<\kappa$ are collapsed, so $\kappa$ is collapsed since it is singular in the extension.