UMP of product functor

Short answer:

1. $(f_1, f_2)$ is just a pair of arrows $f_1$ and $f_2$.
2. $\langle f_1, f_2 \rangle$ is actually just one arrow, and it is the induced one into a product.
3. $f_1 \times f_2$ is also just one arrow, but now between products and it is constructed using projections of the one and the universal mapping property of the other.

Before making this more precise, it might be nice to have an example (which may also explain the notation). Suppose we have sets $A$, $X$ and $Y$, and functions $f_1: A \to X$ and $f_2: A \to Y$. Then we can do the following.

1. Consider just the pair of functions $(f_1, f_2)$.
2. Define a function $A \to X \times Y$, which sends $a$ to the tuple $\langle f_1(a), f_2(a) \rangle$ in $X \times Y$. We denote that function by $\langle f_1, f_2 \rangle$.
3. Define a function $A \times A \to X \times Y$, by sending the tuple $\langle a, b \rangle$ to $\langle f_1(a), f_2(b) \rangle$. We denote this function by $f_1 \times f_2$. So this is saying that we apply $f_1$ to the first coordinate and $f_2$ to the second coordinate.

We can actually formalise all these constructions category-theoretically (assuming our category has products). Similar set-up as above, but now $A$, $X$ and $Y$ are objects in our category and $f_1: A \to X$ and $f_2: A \to Y$ are arrows in our category.

1. Again, just consider a pair of arrows $(f_1, f_2)$.
2. Use the universal mapping property of the product $X \times Y$ to find a unique arrow $A \to X \times Y$ based on $f_1$ and $f_2$. We denote this induced arrow by $\langle f_1, f_2 \rangle$.
3. We can form the product $A \times A$ with projections $\pi_1$ and $\pi_2$ on the first and second coordinate respectively. Then we can form the arrows $f_1 \pi_1: A \times A \to X$ and $f_2 \pi_2: A \times A \to Y$. Then again using the universal mapping property of $X \times Y$ we get a unique arrow $A \times A \to X \times Y$. By the convention explained in point 2, we would denote this arrow by $\langle f_1 \pi_1, f_2 \pi_2 \rangle$, but we then abbreviate this entire construction as $f_1 \times f_2$. Hopefully the example with sets above justifies this notation. So, in other words: $f_1 \times f_2 = \langle f_1 \pi_1, f_2 \pi_2 \rangle$.