Is there a contradiction between "set-theoretic" and "formal" definition of "Big-O"? $O(n) = \{n, n^{2}, n^{1000000}, 2^{n}, ...\}$ [Source A], [Source B]
Say $t_{n} \in O(n)$
By formal definition $t_{n} \leq k \cdot n$  [Source C]


But how can this be? Say $t_{n}$ is actually $n^{2}$, then $t_{n} \leq k \cdot n$ is just false. Would it not be more logical to have $t_{n} \geq k \cdot n$ as a formal definition acompanying $O(n)$?
The first statement you made is not correct. $O(n)$ is the set of all functions $f(n)$ that are bounded above by some constant multiple of $n$ for sufficiently large $n$. The functions you have listed all belong to the set $\Omega(n)$ which is almost like $O(n)$ but instead of being bounded above by $k\cdot n$, this is the set of functions that are bounded below by $k\cdot n$ for sufficiently large $n$.

So, there is no contradiction. But you do need to look at your definitions a bit more carefully.