Let's say you are party $I$, and your friend is party $J$, secret values are $a$ and $b$. The result of the multiplication would look like:
a.b = (a_i \oplus a_j) . (b_i\oplus b_j) = (a_ib_i) \oplus (a_ib_j) \oplus (a_jb_i) \oplus (a_jb_j)
$I$ and $J$ already have these shares and need no communication for $a_ib_i$ or $a_jb_j$. But for $(a_ib_j)$ and $(a_jb_i)$, they do need other's input. $I$ selects a random value r and creates a table with four entries depending on the values of $I$'s and possible values of $J$ with entries $e$ like $r \oplus a_ib_j \oplus a_jb_i$.
a_j = 0,b_j &= 0: e = r \\
a_j = 0,b_j &= 1: e = r \oplus a_i \\
a_j = 1,b_j &= 0: e = r \oplus b_i \\
a_j = 1,b_j &= 1: e = r \oplus a_i \oplus b_i \\
Although these entries look confusing, note that $J$ already knows $a_j$ and $b_j$ values. And any if these entries are $0$, then look at the equation $r \oplus a_ib_j \oplus a_jb_i$. This would have one or two entries evaluate to $0$. That is why the table was constructed that way. Clearly, if $I$ and $J$ engages in a 1-out-of-4 Oblivious Transfer protocol now, then $J$ can only retrieve one entry from the table. So $I$ would be having the random number $r$ and $J$ would be having the computed value xor'ed with $r$. Meaning they share the multiplication result but not the result itself.