Significance of quantifier ordering in quantified boolean formulas (kQBF vs. QBF)
Let me first give a concrete example of a formula $F(x,y)$ such that $\forall x \exists y. F(x,y)$ is true, but $\exists x \forall y. F(x,y)$ is false.
I'll then address the expansion of the universal quantifier in your question.

Take $F(x,y)$ to be a formula expressing the equality $x \oplus y = 1$, where $\oplus$ denotes exclusive-or.
We can explicitly write $F$ as $$F(x,y) = (x \land \neg y) \lor (\neg x \land y).$$

To see that $\forall x \exists y. F(x,y)$ is true, let $x$ be given and take $y = \neg x$.
Then $x \oplus y = x \oplus \neg x = 1$.

To see that $\exists x \forall y. F(x,y)$ is false, first take $x = 1$. Then for $y = 1$, we have that $F(x,y)$ is false, i.e., $F(1,1)$ is false.
By the same reasoning, $F(0,0)$ is false, so there is no $x$ such that $\forall y. F(x,y)$ is true.
Hence $\exists x \forall y. F(x,y)$ is false.



Your mistake is in the following argument.


  Suppose $\exists y\ (F[x \to T] \land F[x \to F])$ is unsatisfiable. Then at least one of $\exists y\ F[x \to T]$ or $\exists y\ F[x \to F]$ is also unsatisfiable.


The point is that $\exists y\ F[x \to T]$ and $\exists y\ F[x \to F]$ may both be satisfiable, but the choice of $y$ which satisfies them may be different.
In order to satisfy $\exists y\ (F[x \to T] \land F[x \to F])$, you must find a single value of $y$ that satisfies both $F[x \to T]$ and $F[x \to F]$ simultaneously.
This may not be possible, even if it is easy to satisfy one of $F[x \to T]$ and $F[x \to F]$ individually.

It may be helpful to expand your argument using the example I gave to see exactly where it breaks down.