A proper definition of $i$, the imaginary unit

The mathematically correct way is to call $i$ a root of $x^{2}+1$, or equivalently to call it a symbol satisfying $i^{2} = -1$. There is no determinacy, nor does there need to be, because $i$ and $-i$ are indistinguishable algebraically (they are both conjugate roots of the same irreducible polynomial in $\mathbb{R}[x]$). There is no "principal" root of $x^{2}+1$.

You need to be careful when you consider $\sqrt{xy}$, obviously; it is only stated to be equal to $\sqrt{x}\sqrt{y}$ when $x$ and $y$ have real square roots.

Finally, complex numbers: sure, algebraic completion of the reals. Perfectly sound definition. A field isomorphic to $\mathbb{R} \times \mathbb{R}$: makes no sense, this is not a field without adding multiplication rules, and adding those is really not motivated unless you ALREADY have $\mathbb{C}$ in mind, which makes it rather circular.

This a definition of a complex number which can resolve many of the problems which you mentioned:

A complex number is a ordered pair of two real numbers: $(a,b),\,\, a,b\in \mathbb R$, with the following definitions of arithmetical operations:

Complex numbers can be added: $(a,b) + (c,d) = (a+c,b+d)$.
They can be multiplied by a real number: $c(a,b)=(ca,cb)\,\,c\in\mathbb R$.
They can also by multiplied: $(a,b) \cdot (c,d) = (ac-bd,ad+bc)$.

Then let the number (1,0) be denoted by 1 and the number (0,1) be denoted by $i$. ¹

It follows from the definition of multiplication that $i\cdot i=(-1,0)=-1$.

With this definition, there is no need for $i$ to be "imaginary" and also it is totally determined that only (0,1) is $i$ and not (0,-1) so it removes the ambiguity of $i=\sqrt{-1}$.

¹ Note: Using this, we have $(a,b) = a(1,0) + b(0,1) = a \times 1 + b \times i$ which can be written as $a+bi$.

To address your final question, the natural extension of the square root function to the complex numbers is done by writing it in polar form, $z=re^{i\theta}$, with $\theta \in [0,2\pi)$, $\sqrt z=\sqrt re^{i \theta/ 2}$. Then if you have a real positive number $A$, $-A=Ae^{i\pi}$, so $\sqrt {-A}=\sqrt Ae^{i \pi /2}=\sqrt Ai$, so no, this is never a problem with positive numbers.

This naturally shows where the problem with breaking apart two square roots occur in the complex numbers: In order to have $\sqrt {z_1z_2}=\sqrt {z_1} \sqrt {z_2}$, we need $Arg(z_1)+Arg(z_2)<2 \pi$, because otherwise we have a problem with the principal arguments changing. (That's an if and only if, btw)