It is sufficient to prove this result for the sum where $i$ ranges from $m 2^n + 1$ to $(m+1)2^n$. That is because $\tilde{H}_k$ is a sum of an odd number of terms of this form with $n= ||k||_2$, so if they all have $2$-adic valuation $2||k||_2$ then $\tilde{H}_k$ does as well.
To that end, consider what happens to the sum if we decrease each $i$ by $2^{n-1}$. That is the same as decreasing the denominator $1/(2i-1)$ by $2^n$.
$1/x - 1/(x-2^n) = 2^n/( x^2)+2^{2n}/( x^3) + \dots$
The terms with a factor of $2^{3n}$ or more are irrelevant, as is the $2^{2n}$ term because the sum over all odd numbers in an interval of length $2^n$ of $1/x$ is clearly even.
Now in $\sum_{i=r}^{r+2^n-1} \frac{2^n}{ (2i-1)^2}$, $1/(2i-1)$ ranges over all odd residue classes modulo $2^{n+1}$, so its square ranges over all residue classes modulo $2^{n+2}$ that are congruent to $1$ mod $8$. The sum of all those residue classes is $2^n + 8 \binom{2^{n-1}}{2} \equiv 2^n$ mod $2^{n+1}$. So the sum is $2^{2n}$ mod $2^{2n+1}$.
After performing this decrease an odd number of times, we are left with $\sum_{i=-2^{n-1}+1}^{i=2^{n-1}} \frac{1}{2i-1}=0$ because the positive and negative terms cancel.
Hence the original sum is $2^{2n}$ mod $2^{2n+1}$.