First, let's focus on $(1)$.
The trick to writing $\phi''$ out is to shift from second-order to first-order thinking, via representatives.
Specifically, "The structure is the disjoint union of three complete graphs" is really a second-order statement as phrased, since it essentially quantifies over sets ("There exist three sets which partition the universe and such that ..."). However, I don't really need to think about the whole "pieces" of the universe: every equivalence class is determined by any of its members.
For example, "There are at least five equivalence classes" can be expressed as "There are at least five elements whose equivalence classes are different," and - granting that we've already written down $\phi'$, so we know that $E$ is an equivalence relation in the first place - this is really just a first-order claim:
$\exists x_1,x_2,x_3,x_4,x_5(\bigwedge_{1\le i<j\le 5}\neg x_iEx_j)$.
Now do you see how to write "There are exactly three equivalence classes" in a first-order way (again, taking $\phi'$ for granted)?
The sentence $$\exists x_1,x_2,x_3[(\bigwedge_{1\le i<j\le 3}\neg x_iEx_j)\wedge\forall y(\bigvee_{1\le k\le 3}yEx_k)]$$ will do the job. I've chosen to use "big-connective" notation here, namely $\bigwedge$ and $\bigvee$, since this emphasizes how the solution generalizes if we replace $3$ with any larger number.
OK, now what about $(2)$?
Well, here the challenge is the same: the statement as written involves quantification over sets, so we need a way to think about representatives somehow.
The obvious notion of "representative" is "member of each 'side,'" i.e., we're looking for a sentence beginning "$\exists x_1,x_2$" where intuitively $x_1$ and $x_2$ lie in the two parts of our bipartite graph. Here the trouble is that we need to figure out a way to identify each "side." So:
Suppose $\mathcal{G}$ is a bipartite graph with domain $U$ and $A,B$ are nonempty sets partitioning $U$ such that $E^\mathcal{G}=(A\times B)\cup(B\times A)$. How can I tell whether two elements of $U$ lie in the same "piece" (= either both in $A$ or both in $B$)?
Now do you see how to use this to solve your problem?
Best Answer
Re: $(i)$, you write:
But that's not true. Remember that Duplicator wins if, at the end of the game, the atomic types of the tuples built are the same. For example, if Spoiler picks $c$, Duplicator picks $4$, Spoiler picks $a$, and Duplicator picks $3$, Duplicator will win the play. We don't check at the end of a play of the Ehrenfeucht-Fraisse game that the tuples built have matching out- (or in-)degrees.
In fact, there is very little Spoiler can do in two moves.