I will put the important point at the beginning. You gave a careful, well-written, and detailed proof. Here it is, copied verbatim:
It is obvious to me that the answer is four:
- Two white socks
- Two black socks
- White on the left foot, black on the right foot
- Black on the left foot, white on the right foot
Perhaps one could omit the personal "to me," it introduces a note of hesitation. The rest is very good, absolutely clear. The "obvious" identifies you as someone with mathematical training.
Then came an attempt, unsuccessful, to manipulate numbers. Such a thing is not a proof, even if one gets the right numerical answer. The idea is primary.
In the analysis that you are (implicitly) making during your calculations, there is no justification for the division $\frac{3}{1}$. Nor is there a justification for addition. You should have written that for every choice of what you put on the left foot, there are $2$ choices of what to put on the right.
Now let us look at a generalization, say to four colours. Suppose that we have $a$ (where a>0) identical pairs of red socks (so $2a$ red), $b$ pairs of yellow, $c$ pairs of green, $d$ pairs of blue. There are "obviously" $4$ choices of colour for the left foot. For every such choice, there are $4$ choices of colour for the right foot, for a total of $(4)(4)$ outfits.
The analysis needs to change if your washing machine eats socks, so that the mates of some socks have vanished. Suppose we have positive numbers $s$, $t$, $u$, $v$, $w$ of red, yellow, green, red, black. If each of $s$, $t$, $u$, $v$, and $w$ is $\ge 2$, there are $5$ choices for the left, and for each such choice there are $5$ choices for the right, total $(5)(5)$.
But if (say) $s$, $t$, and $u$ are $\ge 2$, but $v=w=1$, we need to be careful. True, there are $5$ choices for the left. But it is no longer true that for every such choice, there are $5$ choices for the right. If you put a black sock on the left, there are only $4$ choices for the right.
The general idea still works, but we have to break things up into cases: (i) on the left foot we put a colour that has at least one mate of that colour, and (ii) on the left foot we put a lonely sock.
For case (i), there are $3$ choices for the left, and for each such choice, there are $5$ choices for the right, a total of $(3)(5)$. For case (ii), there are $2$ choices for the left, and for each such choice there are $4$ choices for the right, a total of $(2)(4)$. Finally, add. The total number of choices is $(3)(5)+(2)(4)$.
There are other ways to handle the problem. For example, suppose that we have $m$ colours which each have at least $2$ socks of that colour, and $n$ colours for which we only have a single sock. There are $m$ ways to choose a boring monochromatic pair. For the bicoloured outfits, there are $m+n$ choices of what goes on the left, and for each such choice there are $m+m-1$ choices for the right, a total of $(m+n)(m+n-1)$ bicoloured outfits. The total number of outfits is $m+(m+n)(m+n-1)$. Or else one can count the bicoloured outfits by saying there are $\binom{m+n}{2}$ ways to choose a pair of colours, and for each choice of colours, there $2$ ways to wear them.