For the sake of completeness here is a slightly different approach to part one, which then continues along the path in the link from Brian Scott.
The generating function of the set of partitions by the number of parts is given by
$$ G(z,u) = \prod_{k\ge 1} \frac{1}{1-uz^k}.$$
It follows that the generating function of the set of partitions with an even number of parts is given by
$$ G_1(z) = \frac{1}{2} G(z, 1) + \frac{1}{2} G(z, -1)
= \frac{1}{2} \prod_{k\ge 1} \frac{1}{1-z^k}
+ \frac{1}{2} \prod_{k\ge 1} \frac{1}{1+z^k}.$$
Similarly for an odd number of parts,
$$ G_2(z) = \frac{1}{2} G(z, 1) - \frac{1}{2} G(z, -1)
= \frac{1}{2} \prod_{k\ge 1} \frac{1}{1-z^k}
- \frac{1}{2} \prod_{k\ge 1} \frac{1}{1+z^k}.$$
Therefore
$$ G_1(z) - G_2(z) = Q(z) = \prod_{k\ge 1} \frac{1}{1+z^k}$$
and this is the generating function of the difference between the number of partitions into an even and odd number of parts.
Now observe that this is
$$\prod_{k\ge 1} \frac{1-z^k}{1-z^{2k}}
= \prod_{k\ge 0} (1-z^{2k+1})$$
because the denominator cancels all the factors with even powers in the numerator.
Here is an important observation: the above expression of $Q(z)$ enumerates partitions into unique odd parts in a generating function of signed coefficients where the sign indicates the parity of the number of parts. There is no cancellation between partitions that add up to the same value because the counts of constituent parts have the same parity. Moreover as all parts are odd, partitions of odd numbers must have an odd number of parts and of even numbers, an even number. Therefore the coefficients of $Q(z)$ alternate in sign.
To get the series that generates the absolute values of these coefficients, we create generating functions for the even powers and the odd ones, inverting the sign of the coefficients of the odd ones. The even ones are generated by
$$\frac{1}{2} Q(z) + \frac{1}{2} Q(-z)$$
and the odd ones by
$$-\left(\frac{1}{2} Q(z) - \frac{1}{2} Q(-z)\right)$$
Adding these gives $$ Q(-z).$$
But this is
$$\prod_{k\ge 0} (1-(-z)^{2k+1})
= \prod_{k\ge 0} (1-(-1)^{2k+1} z^{2k+1})
= \prod_{k\ge 0} (1+ z^{2k+1}),$$
precisely the generating function of partitions into distinct odd parts.
This is sequence A000700 from the OEIS.
Best Answer
I'll give a sketch of the bijective proof; ask me if there's some part you don't understand and need fleshed out (or maybe someone else will post a detailed version).
The important idea is that every number can be expressed uniquely as a power of 2 multiplied by an odd number. (Divide the number repeatedly by 2 till you get an odd number.) For instance, $120 = 2^3 \times 15$ and $68 = 2^2 \times 17$ and $81 = 2^0 \times 81$.
Odd parts -> Distinct parts
Suppose you are given a partition of n into odd parts. Count the number of times each odd number occurs: suppose $1$ occurs $a_1$ times, similarly $3$ occurs $a_3$ times, etc. So $n = a_11 + a_33 + a_55 + \cdots$.
Now write each $a_i$ "in binary", i.e., as a sum of distinct powers of two. So you have $n = (2^{b_{11}}+2^{b_{12}}+\cdots)1 + (2^{b_{31}}+2^{b_{32}}+\cdots)3 + \cdots$.
Now just get rid of the brackets, and note that all terms are distinct. (Why?)
E.g. for $20 = 5+3+3+3+1+1+1+1+1+1$, which is a partition of $20$ into odd parts, we do
$20 = (1)5 + (3)3 + (6)1$
$20 = (1)5 + (2+1)3 + (4+2)1$, so you get the partition
$20 = 5 + 6 + 3 + 4 + 2$ in which all parts are distinct.
Distinct parts -> Odd parts
Given a partition into distinct parts, we can write each part as a power of 2 multiplied by an odd number, and collect the coefficients of each odd number, and write the odd number those many times, to get a partition into odd parts.
So for example with $20 = 5+6+3+4+2$ which is a partition of $20$ into distinct parts, we write $20 = 5 + (2)3 + 3 + (4)1 + (2)1$, and then collect coefficients of the odd numbers $5$, $3$ and $1$:
$20 = 5 + (2+1)3 + (4+2)1$
$20 = 5+3+3+3+1+1+1+1+1+1$, which was our original odd partition.
Aside: you asked about the bijective proof, but I cannot resist mentioning that when Euler proved this theorem about partitions, it was with a proof that used generating functions. (When you understand both, maybe you can think about whether this proof is related to the bijective proof.)
Sketch: Let $D(n)$ denote the number of partitions into distinct parts, and $O(n)$ denote the number of partitions into odd parts. Then we have:
$$ \begin{align} \sum_{n\ge0}D(n)x^n &= (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)\cdots \\ &= \frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^6}{1-x^3}\frac{1-x^8}{1-x^4}\frac{1-x^{10}}{1-x^5}\cdots \\ &= \frac{1}{(1-x)(1-x^3)(1-x^5)\cdots} \\ &= (1+x+x^{1+1}+\cdots)(1+x^3+x^{3+3}+\cdots)(1+x^5+x^{5+5}+\cdots) \\ &= \sum_{n\ge0}O(n)x^n \end{align} $$ which proves the theorem.