You mean that functions are formally defined as sets of ordered pairs. Thus, if $I=\{0,1,2\}$ and $X_0=X_1=X_2=\{a,b,c\}$, one member of $\prod\limits_{i\in I}X_i$ is the function $\{\langle 0,a\rangle,\langle 1,b\rangle,\langle 2,a\rangle\}$. You're asking, I take it, how to square this with the idea that a member of $\prod\limits_{i\in I}X_i$ 'ought' to be an ordered triple, for instance $\langle a,b,a\rangle$.
The answer depends on how we choose to define ordered n-tuple. One way, for $n>2$, is to define it as a function whose domain is $\{0,1,\dots,n-1\}$. If we do that, $\{\langle 0,a\rangle,\langle 1,b\rangle,\langle 2,a\rangle\}$ is an ordered triple.
Another approach is to define ordered triples to be ordered pairs of the form $\langle \langle a_0,a_1\rangle,a_2\rangle$, ordered $4$-tuples to be ordered pairs of the form $\langle\langle\langle a_0,a_1\rangle,a_2\rangle,a_3\rangle$, and in general to define ordered $(n+1)$-tuples to be ordered pairs of the form $\langle\pi,a\rangle$, where $\pi$ is an ordered $n$-tuple. If you adopt this definition, then the members of my little product $\prod\limits_{i\in I}X_i$ aren't actually ordered triples. However, there is a natural bijection between them and the 'real' ordered triples, given by $\{\langle 0,x\rangle,\langle 1,y\rangle,\langle 2,z\rangle\}\leftrightarrow\langle\langle x,y\rangle, z\rangle\;,$ by means of which any statement about the one corresponds trivially to a corresponding statement about the other.
Should ordered $n$-tuples be defined in any other reasonable way, a similar situation will obtain.
The function definition may look odd at first, especially when you're dealing only with finite index sets, but it's the only really convenient way to deal with products over infinite index sets.