How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work? 
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the possible values of $0\pmod{2}$ and $1\pmod{3}$?
$$
1\pmod{3} = 1, 4, 7, 10
$$
so on.
Since only $4, 10$ and so on satisfy $\pmod{2}$, only values that satisfy both criteria can be used. 
In general, can we do this for $\pmod{n}$, $n$ being any integer?

Let me try to rephrase what you said. 
Let $10^{10} \equiv a \pmod 6$, $a \in \{0, 1, 2, 3, 4, 5\}$.
As $10^{10} \equiv 0 \pmod 2$, we must have $a \equiv 0 \pmod 2$, so $a \in \{0, 2, 4\}$.
As $10^{10} \equiv 1 \pmod 3$, we must have $a \equiv 1 \pmod 3$, so $a \in \{1, 4\}$.
Therefore, $a = 4$. 
This does give the correct answer as you've noticed. This works because $2, 3 \mid 6$ and $(2, 3) = 1$.
What you are doing is strongly related to the Chinese Remainder Theorem which states that if $n_1, \dots, n_k$ are pairwise coprime positive integers, then for any integers $a_1, \dots, a_k$, the system of congruences $x \equiv a_1 \pmod{n_1},\dots, x \equiv a_k\pmod{n_k}$ has a solution and it is unique modulo $n_1\dots n_k$. In particular, one way to find this solution is to use the method you used (as can be seen here).
