For example, in $\textbf{Z}[\sqrt{10}]$, we have $$6 = 2 \times 3 = (4 - \sqrt{10})(4 + \sqrt{10})$$ and $$10 = 2 \times 5 = (\sqrt{10})^2.$$ How do I use this knowledge to figure out the factorizations of $\langle 6 \rangle$ and $\langle 10 \rangle$? Or is there something else I would need to know before being able to factorize those ideals?
I am aware of $(-1)(2 - \sqrt{10})(2 + \sqrt{10})$, but I'm also aware that it is not distinct from the other factorization I listed above (multiplication by unit). But is that other factorization somehow more "fundamental"? Would it get me quicker to the factorization of $\langle 6 \rangle$?
That's not usually the most helpful unless you know that the element has prime norm, since unique factorization of numbers doesn't hold in general.
Generally you look at norms then try and find ideals above a given ideal. In the case of $(10)$, we know the disciminant of the field is $40$ so $5$ and $2$ ramify, and since the degree of the extension is $2$ that means $5$ and $2$ are each totally ramified. This means
$$(10)=\mathfrak{p}^2\mathfrak{q}^2$$
where $\mathfrak{p}^2=(2)$ and $\mathfrak{q}^2=(5)$ is the unique factorization of $(2)$ and $(5)$. Since $(\sqrt{10})^2=(10)$ and $N(\sqrt{10})=10$ we see that $(\sqrt{10})=\mathfrak{p}\mathfrak{q}$. So
For $(6)=(2)(3)$ we already know $(2)=\mathfrak{p}^2$, so we turn our attention to $(3)$. Reducing $x^2-10$ mod $3$ we get $(x+1)(x-1)$ so $(3)=(3, 1+\sqrt{10})(3,1-\sqrt{10})$ is the factorization of $(3)$, which then gives you the factorization of $(6)$.