In the example you gave, the reason that $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt 2-\sqrt 3)$ is because $\mathbb Q(\pm\sqrt 2+\pm\sqrt 3)$ contains $\sqrt 2$ and $\sqrt 3$:
Notice that $(\pm \sqrt 2+\pm\sqrt 3)^2=2+3\pm 2\sqrt 6$, so $\sqrt 6$ is in the extension, but since $\sqrt 6=\sqrt 3\sqrt 2$, it follows that both $\sqrt 2 $ and $\sqrt 3$ are also in the extension.
This is essentially telling you that $K=\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt 2,\sqrt 3)$, which is the splitting field for $(x^2-2)(x^2-3)$ (which is equivalent to saying that $K$ is a Galois extension), and the fact that $K$ is splitting (Galois) tells us that all the conjugates of any $\alpha\in K$ is also in $K$.
I should add, to answer your question, that what I am essentially saying is that an extension $F(\alpha)/F$ contains all conjugates of $\alpha$ if and only if it is a splitting field (resp. Galois).