I want to find the probability that the first $m$ digits of powers of 2 are a given combination $k_1k_2\dots k_m$. So far, here's my reasoning:
A number $2^n$ will have the first $m$ digits of the form $k_1k_2\dots k_m$ if and only if there exists a non-negative integer $q$ such that
$$ k_1 10^q + k_2 10^{q-1} + \dots + k_m 10^{q-m-1}\leq 2^n <k_1 10^q + k_2 10^{q-1} + \dots + (k_m +1)10^{q-m-1}.$$
Now I'm not sure how to proceed. If it were just the first digit, we could do something like How many times does the most significant digit of $3^n$ is 1?, but in this case I don't know how to generalize the procedure.
The other answer by Sarvesh Ravichandran Iyer you found is more or less the entire answer, except for a simple "trick" (and some extra bookkeeping and probability theory, afterwards). Let the number of decimal digits that $2^n$ has be $s$. If by chance $s$ is a multiple of $m$, we can "reimagine" our base 10 number as a number to the base $10^m$. In this new base, the entire sequence $k_1k_2\dots k_m$ is now a particular value of a single digit, and the answer you found can easily be adapted. In the case where we are less lucky, and $s$ is not a multiple of $m$, we can still proceed by "forcing" our trick by considering $${{2^n}\over{10^{(s\mod m)}}}$$ instead.
More rigorously, let the value of $k_1k_2\dots k_m$ as a decimal number be $r$. Like in the other answer if $2^n$ starts with $k_1k_2\dots k_m$ and has the number of decimal digits $s=tm$, then we know that $$r \times 10^{(t-1)m} < 2^n < (r+1) \times 10^{(t-1)m}.$$ Taking logarithms to the base $10^m$, this tells us that $$t-1 + \log_{10^m} r < n \log_{10^m} 2 < t-1 + \log_{10^m}(r+1)$$
Let $\{x\} = x - \lfloor x\rfloor$ be the distance of $x$ from the first integer below it. The inequality above translates to $$\log_{10^m}r < \{n \log_{10^m} 2\} < \log_{10^m}(r+1).$$ As in the linked answer, $\log_{10^m}2$ is irrational for all positive integers $m$, so the same equidistribution argument applies, and in this case the probability is $$\log_{10^m}(r+1)-\log_{10^m}r.$$ But what happens in the other case, when $u=s\mod m>0$? In that case, we know that $$\begin{array}{c} r \times 10^{(t-1)m+u} < 2^n < (r+1) \times 10^{(t-1)m+u}\\[2ex] r \times 10^{(t-1)m} < {{2^n}{10^{-u}}} < (r+1) \times 10^{(t-1)m} \end{array}$$ for some positive integer $t$, and when we take the logarithm we now get $$t-1 + \log_{10^m} r < n \log_{10^m} 2 -{u\over m} < t-1 + \log_{10^m}(r+1)$$ and $$\log_{10^m}r < \{n \log_{10^m} 2-{u\over m}\} < \log_{10^m}(r+1).$$ It is easy to see that if we subtract any constant from an equidistributed sequence it will remain equidistributed, so in this case the probability remains the same as in the case where $u=0$ and we are spared the necessity of computing the individual probabilities of the various modulo classes (and anyway, again because of equidistribution, they are anyway probably all equal).
In summary, the probability is $$\log_{10^m}(r+1)-\log_{10^m}r.$$