Is the disjunction of these two false statements true?
$$ (\forall x \in \mathbb{N})[x < 3 \implies x \ge 3] \lor (\forall x \in \mathbb{N})[x \ge 3 \implies x < 3] $$ is not true, and your analysis doesn't imply that it ought to be.
What is true by your argument is $$ (\forall x \in \mathbb{N})\bigl[(x < 3 \implies x \ge 3) \lor (x \ge 3 \implies x < 3)\bigr] $$ but you can't move the quantifiers into the middle of its structure and expect it to remain true.
In general, $(A\to B)\lor(B\to A)$ is always true in classical logic, no matter whether $A$ and $B$ are negations of each other or not. Perhaps even more intuition-challenging, $(A\to B)\lor(B\to C)$ is always true too.
Response to edit: I think you're being confused by two different uses of "if ... then" in mathematical English.
On one hand in casual non-formal mathematical English, "if it's heads, then it's tails" would mean something like
In every relevant situation where it's heads, it is also the case that it's tails.
(and one then hopes that the context makes it clear which situations are "relevant").
On the other hand, in (classical) formal logic, the formula "${\rm heads}\Rightarrow{\rm tails}$" means
In the particular situatation we're looking at, it is not the case that it is heads yet it isn't tails.
Which is the same as saying
The particular situation we're looking at is not a counterexample to sense (1).
Confusion creeps in because "${\rm heads}\Rightarrow{\rm tails}$" is often pronounced "if heads then tails", as if its meaning were the same as in (1), but it is not when formulas are involved.
The sense in which you can prove that $(A\Rightarrow \neg A)\lor(\neg A\Rightarrow A)$ is true is (2) and only (2). It stops being true if you try to reinterpret the $\Rightarrow$s as if they meant (1).
Furthermore it it looks like you're also confusing yourself by the difference between "is true" and "has a proof" when you say that "this or that" is true but "this" and "that" are both false.
What you mean here is that you believe you have a proof of "this or that". But clearly it is not true because the definition of "or" is that "this or that" is true exactly when at least one of "this" and "that" is true -- and you have just argued that they are both false.
So when you ask,
How can "this" be false and "that" be false and "this or that" be true?
the answer is that they can't and you're mistaken about at least one of those parts. Same as if you have an apparent proof that concludes $2+2=3$: you shouldn't be asking "why is $2+2$ not $4$?" but "what is wrong with this thing that looks like a proof but obviously can't be?"
For any arbitrary $x$ you have the schema: $\phi(x)\vee\psi(x)$
$$(A(x)\to\neg A(x))~\vee~(\neg A(x)\to A(x))$$
This does not suggest that: $(\forall x~\phi(x))~\vee~(\forall x~\psi(x))$.
What the schema means is that: $\forall x~(\phi(x)\vee\psi(x))$ .
$$\forall x~[(A(x)\to\neg A(x))\vee(\neg A(x)\to A(x))]$$
In breif; the schema must always be applied to the same entity. The universal quantfier will not distribute over disjunction.
Then the statement, "if the coin shows heads, then it shows tails, or if the coin shows tails, then it shows heads" is true, even though both disjuncts are false.
Oh, no, you will find that they are not both false; one will always be true. Let us look closer.
Okay, I just flipped a coin and it shows tails. I can thereby truthfully asset that "If this coin shows heads, then it shows tails." That material conditional statement is held to be true (since the antecedant is false).
Okay, I just flipped several more coins and now have one showing heads. I can now truthfully asset that "If this coin shows tails, then it shows heads," because now that material conditional statement is the one held to be true.
So whatever coin toss I have, if it shows heads or tails, then one from the two conditional statements will hold true. (Indeed, only one will hold true for any coin toss.)
Perhaps what is counter-intuitive is the fact that whenever $A$ is false, then $A\Rightarrow B$ is true, based on how logical implication is defined: $$ \begin{array}{|c|c|c|} \hline A & B & A\Rightarrow B \\ \hline {\bf F} & F & {\bf T}\\ {\bf F} & T & {\bf T}\\ T & F & F\\ T & T & T\\ \hline \end{array} $$
Applied to the example of the coin:
if the coin shows heads, then it shows tails
or
if the coin shows tails, then it shows heads
Let's say a coin is flipped, and it shows heads.
Then the following statement is false:
the coin shows tails
and thus the second half of the disjunction is true:
if the coin shows tails, then it shows heads
therefore the whole disjunction is true.
Similarly if tails were shown.