Version 1 does not work. For consider the order $\leq'$ defined by
$$a \leq' b :\equiv \begin{cases}
b \leq a & a, b < 0 \\
a \leq b & otherwise
\end{cases}$$
This is a total order on $\mathbb{R}$. Furthermore, one can show that $(\mathbb{R}, \leq')$ and $(\mathbb{R}, \leq)$ are order-isomorphic, so $(\mathbb{R}, \leq')$ also satisfies any other properties of the order of $\mathbb{R}$ which are phrased purely in terms of the order relation (and not in reference to any other functions or operations defined on $\mathbb{R}$).
$(\mathbb{R}, \leq')$ satisfies Version 1 of the axioms, but it clearly fails to satisfy Version 2.
In fact, consider any total order $\leq''$ on $\mathbb{R}^2$ which satisfies Version 2. Note that version 2 implies that $0 \leq'' c^2$ for any $c$. First, note that this follows immediately for $0 \leq c$, since we would have $0 \leq c \cdot c$. And for $c \leq 0$, we see that $-c + c \leq -c + 0$, so $0 \leq -c$, so $c^2 = (-c)^2 \leq 0$.
Now suppose that $a \leq b$. Then there exists $c$ such that $a + c^2 = b$. Then we see that $a + c^2 + 0 \leq'' b + c^2$. Then $a + c^2 \leq'' b + c^2$. Adding $-(c^2)$ to both sides gives us $a \leq'' b$.
So therefore, we see that $\leq''$ agrees with $\leq$.
Therefore, version 2 is enough to completely specify the order relation on $\mathbb{R}$.
In fact, consider any ordered ring $(R, \leq)$ in which all nonnegative elements have square roots. Then version 2 uniquely specifies the order on $R$ by the above argument.
Note that in rings where not every nonnegative element has a square root, there can be more than one total order satisfying (2). For consider $R = \{a + b \sqrt{2} \mid a, b \in \mathbb{Q}\}$, which is an ordered field under the normal $\leq$ order inherited from $\mathbb{R}$. But there is a nontrivial ring isomorphism $R \to R$, given by $a + b\sqrt{2} \mapsto a - b\sqrt{2}$. This indicates there must be another order relation $\leq'$ on $R$, defined by
$$a + b \sqrt{2} \leq' c + d \sqrt{2} :\equiv a - b \sqrt{2} \leq c - d \sqrt{2}$$
which also satisfies the ordered ring requirements. This suffices to prove that $R$ is not closed under the taking of square roots of nonnegative elements.
