Understanding non-truths and writing them as contradictions A contradiction is defined as a logical proposition that is always false, such as
$(p \land \neg p) \iff False $
According to my Professor, examples of non-truth are
(1) Assuming that $(p\rightarrow q)$ and $(\neg p \rightarrow \neg q )$ are logically equivalent.
(2) $(\exists u$ $p(u))$ $\land$ $(\exists u$ $q(u))$ $\rightarrow$ $(\exists u$ $(p(u) \land q(u)))$.
So from what I understand, a non-truth is when we make a logical mistake (state a logical equivalence that doesn't exist).
Could we write (1) and (2) this way:
$((p\rightarrow q) \iff (\neg p \rightarrow \neg q )) \iff False$
$[((\exists u$ $p(u))$ $\land$ $(\exists u$ $q(u))$ $\rightarrow$ $(\exists u$ $(p(u) \land q(u)))) \iff (a \rightarrow b)]$ $\iff False$.
edit: I wrote a typo, I meant $(p \land \neg p) \iff False $, not $(p \land \neg q) \iff False $. Very sorry about that.
 A: Here (1) and (2) are 'non-truths' because they aren't always true - you can find propositions $p,q$ such that (1) or (2) doesn't hold.
For example if we set
$p=$"$P$ is a rectangle"
$q=$"$P$ is a square", then:
$p \rightarrow q$
doesn't hold, for not all rectangles are squares
$\neg p \rightarrow \neg q$
does hold, because what isn't a rectangle cannot be a square.
We found $p,q$ such that
$$(p\rightarrow q) \iff (\neg p \rightarrow \neg q)$$ does not hold so it's not a tautology - 'truth'. We call such sentences 'non-truths'.
You may convince yourself that the following propositions show, that (2) is a 'non-truth', for the same reason.
$p=$"$P$ is a rectangle with area $u$",
$q=$"$P$ is a square with area $u$".
Now, 'non-truth' isn't necessarily false.
If we take $p=q$ then both (1) and (2) evaluate to true. Basically:
-a tautology is always true
-a falsehood is always false
-a 'non-truth' can go either way
A: 
A contradiction is defined as a logical proposition that is always false.

The keyword is "always".  It means "all truth values we may give to its literals/predicates".
$p\land\neg q$ is not always false.   We may value it as true by valuing $p$ as true and $q$ as false.
$p\land\lnot p$ is always false, because whatever truth value we give to $p$, we will evaluate $p\land\lnot p$ as false.
A: 
A contradiction is defined as a logical proposition that is always
false,

The logical proposition ‘I am a swan’ is arguably “always” false, yet it is not a contradiction, because its truth-functional form $P$ is sometimes true.

such as $$(p \land \neg q) \iff\text{False}.$$

We can fully symbolise the above as $$(p \land \neg q) \iff \bot,\tag0$$ where $\bot$ is the standard symbol for contradiction, i.e., $(p\land\lnot p).$
Now, put $(p,q)=(T,F)$ to see that $(p \land \neg q)$ isn't a contradiction. So, $(0)$ is a false assertion.

According to my Professor, examples of non-truth are
(1) assuming that $(p\rightarrow q)$ and $(\neg p \rightarrow \neg q)$ are logically equivalent

$$(p\rightarrow q)\leftrightarrow(\neg p \rightarrow \neg q)\tag1$$

(2) $$(\exists u p(u)) \land (\exists u q(u)) \rightarrow (\exists u (p(u) \land q(u)))\tag2$$

Sentences $(1)$ and $(2)$ are indeed invalid (what you mean by them being “non-truths”).
Now, put $(p,q)=(T,T)$ to make sentence $(1)$ true!
Next, put $p(u),q(u)=\text‘u$ is a member of the universe’ to make sentence $(2)$ true!
We thus say that these two sentences are satisfiable.

Can we write $$[((\exists u p(u)) \land (\exists u q(u)) \rightarrow (\exists u (p(u) \land q(u)))) \iff (a \rightarrow b)] \iff \text{False}$$

$$\big(\exists u p(u) \land \exists u q(u) \rightarrow \exists u p(u) \land q(u) \big) \leftrightarrow (a \rightarrow b)$$ is neither a contradiction nor even unsatisfiable.
Incidentally, the sentence $(\exists x\:x\ne x)$ is unsatisfiable but not a propositional-logic contradiction.

Understanding non-truths and writing them as contradictions

Summing up: an invalid sentence might not be a contradiction nor even unsatisfiable.
I appreciate that all this can be rather confusing at first; this, this and this (and the links within them) could be further reading.
