You've answered your own question. You say you're familiar with the fact that $(-b)c = -bc$ and $b(-d) = -bd$ in a ring. Now, if you write down a definition of multiplication that satisfies all of those properties, you get a ring structure on the set of equivalence classes, and $(-b)c = -bc$ and $b(-d) = -bd$ in a ring...!
Perhaps some extra generality will make this argument seem less circular. You have a rig $R$. This rig has an underlying additive monoid $M = (R, +)$. This abelian monoid has a Grothendieck group $G(M)$, which is the universal group into which $M$ maps. More precisely, there is a forgetful functor $F : \text{Ab} \to \text{CMon}$ from the category of abelian groups to the category of commutative monoids, and the Grothendieck group functor is its left adjoint. $G(M)$ consists of equivalence classes of formal expressions $m - n$ where $m, n \in M$ with addition defined as expected.
Now you want to put a ring structure on $G(M)$ compatible with the multiplication on $R$. It turns out that what you are actually constructing is the universal ring into which $R$ maps; more precisely, there is a forgetful functor from the category of rings to the category of rigs and you are constructing its left adjoint. Well, in any such ring we need
$$(a - b)(c - d) = ac - bc - ad + bd$$
as you well know, so this is the unique possible ring structure on $G(M)$ compatible with the multiplication on $R$ (and distributivity, inverses, etc.) if it is well-defined. This is a general property of adjoint functors (they are unique up to unique isomorphism if they exist). The only thing left is to verify that it actually works.
Another way to say the above is the following. Let $R$ be a rig and let $S$ be a ring, and let $\phi : R \to S$ be any rig homomorphism whatsoever. Then
$$(\phi(a) - \phi(b))(\phi(c) - \phi(d)) = \phi(ac) - \phi(bc) - \phi(ad) + \phi(bd).$$
There is nothing circular about this because $S$ is a ring and the above computation takes place in $S$.
Travelling down the alternative road, we see that
\begin{align*}
\sqrt{x+a} &= \sqrt{y+b} & \\
\sqrt{x+a}\times\sqrt{y+b} &= \sqrt{y+b}\times\sqrt{y+b} &(\text{multiplying both sides by}\ \sqrt{y+b})\\
\sqrt{x+a}\times\sqrt{y+b} &= y + b &(\text{simplifying})\\
\sqrt{x+a}\times\sqrt{x+a} &= y + b &(\text{using the equation}\ \sqrt{x+a}=\sqrt{y+b})\\
x + a &= y + b &(\text{simplifying}).
\end{align*}
Multiplying by $\sqrt{x+a}$ has exactly the same effect on the equation as multiplying by $\sqrt{y+b}$, but we can replace one expression by the other whenever it is convenient. For example, going from the third equation to the fourth, $\sqrt{y+b}$ on the left hand side was replaced with $\sqrt{x+a}$; this was done to allow for algebraic simplification.
Best Answer
Your work is just fine: you've shown you know that $\dfrac 12 = \sqrt{\dfrac{1}{4}},\;\;$ and that for any $x,y\in \mathbb{R^+\cup \{0\}},\;\;\sqrt x \cdot \sqrt y=\sqrt{xy}$.
The negative sign outside of the radicand has no impact on your operations: since the operations between terms is strictly multiplication, we can operate (multiply) as if the positive terms are entirely contained within parentheses, all of which is then multiplied by $-1$:
$$-\frac{1}{2}\cdot \sqrt{\frac{2}{5}} =-\left(\frac{1}{2}\cdot \sqrt{\frac{2}{5}}\right) = -\left(\sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{5}}\right)$$ $$= -\left(\sqrt{\frac{1\cdot2}{4\cdot 5}}\right)=-\left(\sqrt{\frac{1}{10}}\right) = -\sqrt{\frac{1}{10}}$$
Note: the parentheses are used for illustration only: to make explicit that your computation is indeed correct. But, in fact, parentheses are not necessary.