Q:

What does it mean "adjoin A to B"?

What does it mean that we can obtain $\mathbb{C}$ from $\mathbb{R}$ by adjoining $i$?
Or that we can also adjoin $\sqrt{2}$ to $\mathbb{Q}$ to get $\mathbb{Q}(\sqrt{2})=\{a+b \sqrt{2}\mid a,b \in \mathbb{Q}\}$? 

A:

"Adjoin" generally means "find the smallest (field / ring / favorite object) extension containing the object that we're adjoining." So the very smallest field that contains both $\mathbb{R}$ and $i$ is in fact $\mathbb{C}$, and $\mathbb{Q}[\sqrt{2}]$ denotes the smallest field extension of $\mathbb{Q}$ containing $\sqrt{2}$.

A:

It means exactly what you have written there. Let $F$ be a field and $\alpha$ be a root of a polynomial $f(x)$ that is irreducible of degree $d$ over $F[x]$. Then we say we can adjoin $\alpha$ to $F$ by considering all linear combinations of field elements of $F$ with scalar multiples of powers of $\alpha$ up to $d-1$, or rather:
$F(\alpha) = \{a_{1} + a_{2}\alpha + a_{3} \alpha^{2} + \cdots + a_{d}\alpha^{d-1}| a_{1},\ldots,a_{d} \in F\}$. 
Note that because $\alpha \notin F$ because it is a root of an irreducible polynomial in $F[x]$. However, $F(\alpha)$ certainly contains $\alpha$. In this light, we can view field extensions as a way of "extending" base fields to include elements they wouldn't otherwise have. 
You'll notice this is consistent with the definition of $\mathbb{C}$ from $\mathbb{R}$. $i$ is the root of the irreducible polynomial $x^{2} + 1$ over $\mathbb{R}[x]$. Hence, if we define $\mathbb{C} = \mathbb{R}(i)$, then 
$\mathbb{C} = \{a + bi | a, b \in \mathbb{R}\}$. 

