Show that if $\langle f,g\rangle=\langle k,h\rangle$ then $f=k$ and $g=h$

Briefly, you deduced correctly the equalities $pr_a \circ \langle k,h\rangle=k$ and $pr_a \circ \langle k,h \rangle =f$, right? Hence $k=pr_a \circ \langle k,h \rangle =f$. And as you said the same works for $pr_b$.

Observe that a stronger property holds: $\langle f,g \rangle =\langle k,h \rangle$ if and only if $f=k$ and $g=h$. This is precisely the definition of product you wrote.

The notation $a \times b$ can be confusing when you're first learning about products in categorical terms. $a \times b$ is simply an object that happens to satisfy the universal property of the product. Depending on the underlying category, it may or may not be the cartesian product. Similarly, $\langle f, g\rangle$ is the arrow induced by the universal property. It's certainly not an ordered pair.

To illustrate, we can rewrite the definition of the product without using any suggestive notation. The product of two objects $a$ and $b$ in a category $\mathcal C$ is an object $d$ with a pair of arrows $p_a : d \to a$ and $p_b : d \to b$ such that for any object $c$ and arrows $f : c \to a$ and $g : c \to b$, there exists a unique arrow $u : c \to d$ satisfying $f = p_a \circ u$ and $g = p_b \circ u$.

Your proof of the question given is exactly right. Since $\langle f, g\rangle = \langle k, h\rangle$, we have $$ f = p_a \circ \langle f, g\rangle = p_a \circ \langle k, h\rangle = k. $$

And similarly, $g = h$.