There are a couple of things at play here. One is the fact that for each partition $\lambda$ we have
$$\sum_{(i,j)\in\lambda}(-1)^{i+j}\cdot h_{ij}(\lambda)=\frac{(\lambda_1-\lambda_2+\lambda_3+\cdots)+(\lambda'_1-\lambda'_2+\lambda'_3+\cdots)}{2}.$$
This is easy to see from the description I mentioned in the answer to your previous question: this counts the number of cells with both coordinates even minus the cells with both coordinates odd. Now if you pair each partition with its conjugate, you see that summing this statistic over all partitions of $n$ is equivalent to simply summing the statistic $\lambda_1-\lambda_2+\lambda_3+\cdots$. This is as bijective as this argument will get because these two statistics aren't equally distributed among partitions of $n$.

Now it turns out that this last statistic $\lambda_1-\lambda_2+\lambda_3+\cdots$ *is* equally distributed to the statistic "number of odd parts", and this can be proven bijectively. In fact this is essentially exercise 3.4.5 in Igor Pak's "Partition Bijections: A Survey" (you would have to combine it with Franklin's bijection as described in 3.3.1 to get the full result).