big-O and Θ notation subset
Lets refactor and reword these statements for ease of thought.

Let $A(n)$ be $Θ(g(n))$.

Let $B(n)$ be $O(g(n))$.


  Note that $A$ implies $B$ because $A$ is stronger than $B$.


This means for $A$ to be fulfilled we have to fulfill the criteria of $B$ and more. Therefore $A$ being fulfilled implies $B$ must also be.


  Written set theoretically, we have the set of functions that fulfill the criteria of $A$ are a subset of the functions that fulfill the criteria of $B$.


Well this is not contradictory, but actually equivalent in a sense to the last statement which showed that $A$ has the criteria of $B$ and more. 

From our understanding of this point we can determine that: All the functions that fulfill the criteria of $A$ fulfill $B$, but not all functions that fulfill the criteria of $B$ fulfill the criteria of $A$.

This is really an equivalent statement to: The functions that fulfill the criteria of $A$ are a subset of the functions that fulfill the criteria of $B$ because the second set contains the first.

And therefore $\Theta(g(n))\subseteq O (g(n))$ is true, and it is equally true that $f(n) = \Theta(g(n))$ implies $f(n) = O(g(n))$ without any contradiction.

Furthermore this makes sense with the definition for $\Theta(g(n))$ which states that:
$f(n)\in \Theta(g(n))$ if and only if $f(n) \in O(g(n))$ and $f(n) \in \Omega(g(n))$. Clearly by this definition, $f(n) \in Θ(g(n))$ does imply $f(n) \in O(g(n))$, and $\Theta(g(n))$ is clearly a subset of $O(g(n))$ as well.