If I understand the question correctly, brute-force search with some Julia code gives following small values $C(m,n)$ for the maximum number of subsets covered by the contiguous substrings:
\begin{array}{rr|rrrrrrrrrrrrrrrrr}
&& n\\
&& 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\
\hline
m & 2 & 4 & = \\
& 3 & & 7 & 8 & = \\
& 4 & & & 11 & 13 & 14 & 15 & 16 & = \\
& 5 & & & & 16 & 19 & 21 & 23 & 25 & 27 & 30 & 31 & 32 & = \\
& 6 & & & & & 22 & 26 & 29 & 32 & 35 & 38 & 41 & 45 & 48 \\
& 7 & & & & & & 29 & 34 & 38 & 42 & 46 & 50 & 54 & 58 \\
& 8 & & & & & & & 37 & 43 & 48 & 53 & 58 & 63 & 68 \\
& 9 & & & & & & & & 46 & 53 & 59 & 65 & 71 & 77 \\
& 10& & & & & & & & & 56 & 64 & 71 & 78 & 85 \\
& 11& & & & & & & & & & 67 & 76 & 84 & 92 \\
& 12& & & & & & & & & & & 79 & 89 & 98 \\
& 13& & & & & & & & & & & & 92 & 103 \\
& 14& & & & & & & & & & & & & 106 \\
\end{array}
"=" means that the rest of the row repeats the last value, that is, there is a string that exhibits all $2^m$ subsets (adding more length will not change this).
Looking at diagonals:
* The main diagonal is as described in the question.
* The first superdiagonal 4,8,13,19,26,34,43,53,64,76 has [one OEIS match][1], A034856 "a(n) = binomial(n+1, 2) + n - 1 = n*(n + 3)/2 - 1".
* The second superdiagonal has no OEIS entry but OEIS [guesses][2] it to be a quadratic. Same with third superdiagonal.
* The fourth superdiagonal, starting from 16,25,35,46,58,71,85,100,116,133,151 has [two matches][3] (but different values before the 16).
If we write $T(m)$ for the smallest $n$ where $2^m$ appears, $T$ forms a sequence that starts 1,2,4,8,13, which matches many things in OEIS. Examples of shortest strings where all $2^m$ subsets of alphabet $\{1,2,\ldots,m\}$ appear are:
\begin{array}{l|l}
m & T(m) \\
\hline
1 & 1 & 1 \\
2 & 2 & 12 \\
3 & 4 & 1231 \\
4 & 8 & 12314234 \\
5 & 13 & 1234512413524 \\
\end{array}
Proof sketch for first superdiagonal
------------------------------------
From the small values one can conjecture that $C(m,m+1) = \frac{1}{2}m^2 +
\frac{3}{2}m - 1$. Indeed this seems to be the case.
Let us construct the string letter by letter. At each position we have
two kinds of choices: a letter already seen, or an unseen letter. In
the second case we can, without loss of generality, pick the *smallest
unseen letter*; this is just a matter of permutation of the alphabet.
In the first case, it does no good to pick the previous letter again,
so we rule that out.
Consider a string over alphabet $\{1,2,\ldots,m\}$ of length $n=m+1$.
Because each letter has to appear at least once, we have only one
letter $q$ appearing twice, with the repeat occurring at some position $p$.
So we have only a small number of strings to consider. The strings are of the form:
$$
1,2,\ldots,p-1,q,p,\ldots,m
$$
where at position $p$ we have picked an already seen letter $q \in
\{1,2,\ldots,p-2\}$. The rest of the string is forced because we must pick all the remaining $m-p+1$ letters, and w.l.o.g. we can just take them in order.
For example, with $m=5$ and $n=6$, our candidate strings and the
numbers of subsets covered are as follows. The repeated letter
$q$ is underlined.
\begin{array}{ll}
12\underline{1}345 & 16 \\
123\underline{1}45 & 17 \\
123\underline{2}45 & 16 \\
1234\underline{1}5 & 18 \\
1234\underline{2}5 & 17 \\
1234\underline{3}5 & 16 \\
12345\underline{1} & 19 \\
12345\underline{2} & 18 \\
12345\underline{3} & 17 \\
12345\underline{4} & 16
\end{array}
Taking $p=n$ and $q=1$ we get the string that has first all letters $1,\ldots,m$ in order, then repeats the 1. This gives the conjectured value. I believe that for other choices of $p$ and $q$ one can show that the result is smaller (for all $m$) but I didn't check the details.
The $m=6$ case
--------------
**Update (25.5.2021).** We have $C(6,24)=64$. There are several 24-digit strings attaining this maximum, for example $$123456 \; 532614 \; 251364 \; 265143$$ (spaces just for easier reading). This string begins with 1...m, supporting Per Alexandersson's conjecture in the comments.
It also seems (by about 56 cpu-hours of computation) that $C(6,23)=63$, so the $T$ sequence would begin $1,2,4,8,13,24$. This has [7 matches in OEIS][4], but I could not easily see a connection to the present problem in any of them.
[1]: https://oeis.org/search?q=4%2C8%2C13%2C19%2C26%2C34%2C43%2C53%2C64%2C76&sort=&language=&go=Search
[2]: https://oeis.org/search?q=8%2C14%2C21%2C29%2C38%2C48%2C59%2C71%2C84&sort=&language=&go=Search
[3]: https://oeis.org/search?q=16%2C25%2C35%2C46%2C58%2C71%2C85%2C100%2C116%2C133%2C151&sort=&language=&go=Search
[4]: https://oeis.org/search?q=1%2C2%2C4%2C8%2C13%2C24&go=Search