Since I find your notation bizarre, I hope you don't mind if I rewrite your Question 1 in more standard notation.
Fix a language $L$. Let $I$ be a non-empty set, and let $(\varphi_i)_{i\in I}$ be a family of $L$-formulas, all of which have free variables from a set $V$ of variables. For each $i\in I$, let $V_i\subseteq V$ be the finite set of variables which are free in $\varphi_i$. We also pick a variable $x_i\in V$ for each $i\in I$, which may or may not be free in $\varphi_i$.
Let $M$ be an $L$-structure. An assignment $a = (a_x)_{x\in V}$ is a family of elements from $M$ indexed by $V$ (so the variable $x$ gets assigned to $a_x\in M$). Given assignments $a$ and $b$, we write $a\equiv_i b$ if $a_x = b_x$ for all $x\in (V_i\setminus \{x_i\})$.
Fix an assignment $a$ and an ultrafilter $U$ on $I$.
Statement 1: For all assignments $b$, if $\{ i \mid a \equiv_i b \} \in U$, then $\{ i \mid M\models \varphi_i(b) \} \in U. $
Statement 2: $ \{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} \in U.$
Are these two statements equivalent?
No, they are not equivalent. To make their logical structure more transparent, let's rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.
Statement 1: $\forall b\, ((\forall^* i\, a\equiv_i b)\rightarrow (\forall ^* i\, M\models \varphi_i(b)))$
Statement 2: $\forall^* i \forall b\, (a\equiv_i b\rightarrow M\models \varphi_i(b))$
Now the nice thing about the $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.
Unfortunately, the quantifier $\forall^*$ does not commute with $\forall$. Statement 2 implies Statement 1, since $\forall^* i\forall b$ is stronger than $\forall b\forall^* i$. But the converse does not hold.