Suppose that $\exists x\in X$ $\forall y \in X \exists g\in G \ : g(x)=y $.
Now let $y_1$, $y_2 \in X$. Choose $g_1$, $g_2$ such that $g_1(x)=y_1$ and $g_2(x)=y_2$ which you can do by original assumption. Set $g=g_2 g_1 ^ {-1}$. Clearly $g(y_1)=y_2$.
Therefore the two definitions of transitivity are in fact equivalent, even though one of them seems weaker at first.
The way to think about transitivity is that by acting with a group, you can get anywhere in your set from any starting point. Orbits of the group are precisely sets of points that you can reach with your group action from some starting point. Hence transitivity means that there is only one orbit. For example if you consider group of rotations with respect to fixed point acting on Euclidean space, orbits are just spheres of constant radius with respect to that point. Group action restricted to these spheres is transitive.
For the first one:
You just have to show, that $gx \neq y$ for all $g \in G_y$ and $x \in X \setminus \{y\}$ since all group action axioms are already fullfilled from $G_y \subseteq G$. I hope you can do this (try it! (really!) If you get stuck, post a comment).
Towards the second one:
First consider that you have always at least 2 orbits, since the orbit of $(x,x) \in X \times X$ has always(!) the same two elements in each component (make this yourself clear) and since you have two different elements $x,y \in X$ the element $(x,y)$ is not in this orbit.
For one direction you have to show two orbits are enough, if every $G_y$ is transitive on $X \setminus \{y\}$:
First show, that $G$ is transitive on $X$ (meaning there is only one orbit). Here it is sufficient, that $G$ does not have a fixpoint $y$ (because then there exists $g \in G$ with $gy\neq y$ and because $G_y$ is transitive on $X \setminus \{y\}$ the orbit of $y$ is already everything). Because you have at least 3 elements $x,y,z$ and each element is contained in $X \setminus \{p\}$ for some $p$ you can do this. But this shows, that $\{(x,x) \ | \ x \in X\}$ is already one orbit.
Now you have to show, that $\{(x,y) \ | \ x,y \in X, x \neq y \}$ is an orbit. For fixed $y$ and since $G_y$ acts transitive you can see, that $\{(z,y) \ | \ z \in X, z \neq y\}$ is in the orbit of $(x,y)$ for any $y \neq x \in X$. And samewise for fixed $x \in X \{(x,z) \ | \ z \in X z \neq x\}$is contained in the orbit of $(x,y)$ for any $x \neq y \in X$. Since $x,y$ where arbitrary you can deduce that $\{(x,y) \ | \ x,y \in X, x \neq y \}$ lies in an orbit of $(x,y)$ and since these are now every elements of $X \times X$ you are done.
For the direction backwards consider some $G_y$ which is not transitive and the orbit of $(x,y)$ for some $x \in X$. Now show, that you don't get all elements of this form and hence you need $\geq 3$ different orbits.
Best Answer
i. It is redundant to state that the action is transitive - showing that the action is transitive using the definition above is immediate by choosing $x=y', y=x'$. It is however a necessary condition for an action to be doubly transitive.
ii. Using your definition of doubly transitive, the proposition, and (i), you can restate the definition of doubly transitive to the one you wrote, but don't forget to also ask for $X$ to be of size at least 2.
Edit - There's a small gap in my answer for (ii) - When you don't require the action to be transitive, it's not clear that the diagonal is even an orbit. The answer is correct anyway, since if the diagonal is not an orbit, then its complement is also not an orbit (see David's comment).