With the given notation explain how any function $f: A \to B$ determines a section of $\pi_A$ 
$\pi_A ((a, b)) := a$ and $\pi_B((a, b)) := b$.

I think a section is a right-inverse of a function. So, $\pi_A \circ f(a) = \pi_A(f(a)) = \pi_A(b) = \pi_A(\pi_B((a, b))) = a.$
I am not sure if any of this makes sense seeing as how $\pi_A$ takes a pair, not just a single element from either $A$ or $B$. Am I at least goig in the right direction?
 A: You're correct about what a section is.
You're also right to be unsure if your equation makes any sense: indeed, because $\pi_A$ takes a pair (i.e. an element of $A\times B$), the expression $\pi_A(f(a))$ is meaningless as $f(a)$ is in $B$, not $A\times B$.
You need to use $f$ to construct a section of $\pi_A$, but $f$ itself is not a section. A section is going to be a function from $A$ to $A\times B$ (i.e. from elements of $A$ to pairs), not from $A$ to $B$. Your job is to figure out the right function.
One other hint: since $f$ is a completely arbitrary function, it doesn't have to be injective, in other words it might "forget" some information about its input, e.g. if it maps all elements of $A$ to a single element in $B$, then its output has completely forgotten its input. On the other hand, a section of $\pi_A$ must remember its input (i.e. it must be injective). Otherwise, how could the original input be recovered by the application of $\pi_A$?
So, you need to construct a function from $A$ to $A\times B$ that uses $f$ in some way but also completely remembers its input.
