Starting where you left off:
$S(m)=S(m/2)+ \Theta(\lg m)$
Compared this to the generic recurrence:
$T(n) = aT(n/b) + f(n)$
Let's address the questions, you raised.
What does "polynomially" larger mean?
A function $f(n)$ is polynomially larger than another function $g(n)$, if $f(n) = n^i g(n)/t(n)$, where $t(n)$ is some sub-polynomial factor such as $\log n$, etc that appears in $g(n)$. In other words, we want to see, ignoring sub-polynomial and constant factors, if $f(n)$ is a polynomial multiple of $g(n)$.
Does case 2 apply here?
Recall that Case 2 of the master theorem applies when $f(n)$ is roughly proportional to $n^{\log_b a}\log^kn$ for some $k \ge 0$. Now, applying all this to your equation above, $a = 1$, $b=2$ and $f(m) = \Theta(\log m)$. This give $k = 1$ and since $m^{\log_2 2^0}\log^km = \Theta(\log m)$, case 2 does apply.
What is the solution using Case 2?
Generally, when case 2 does apply, the solution is always of the form $\Theta(n^{\log_b a}\log^{k+1}n)$. So, in your case, it'll be $\Theta(\log^2 m)$ as you've already figured.
There are some missteps in your application of the master theorem.
First
$$
T(n) = 7\left[ 7T\left(\frac{n}{4}\right) + \frac{1}{4}n^2\right] + n^2 = 49 T\left(\frac{n}{4}\right) + \frac{11}{4}n^2,
$$
so let's call $a = 49$, $b = 4$, and $f(n) = \frac{11}{4}n^2$. Clearly $f(n) \in O(n^2)$, so let's also call $c = 2$. Then
$$
2 = c < \log_b a = \log_4 49 \approx 2.81,
$$
so we can apply case 1 of the master theorem (according to wikipedia) to find that
$$
T(n) \in \Theta\left(n^{\log_4 49}\right).
$$
Further this form of the recurrence allows us to see immediately that $T(n) \geq T'(n)$ if $a \leq 49$, presuming the initial values are the same. (Why? This is important.) $\tag{$\dagger$}$
Now let's look at the recurrence for $T'$, which is
$$
T'(n) = a T'\left(\frac{n}{4}\right) + n^2.
$$
We'll re-use the variable names from before, but in this case set $a = a$, $b = 4$, and $f(n) = n^2$. Again $f(n) \in O(n^2)$, so $c = 2$. If $a > 16$ then $2 = c < \log_4 a$, so case 1 of the master theorem again tells us that
$$
T'(n) \in \Theta\left(n^{\log_4 a}\right).
$$
Of course
$$
n^{\log_4 49} \in o\left(n^{\log_4 a}\right)
$$
if $a > 49$, so we conclude that $T(n) \in o\Bigl(T'(n)\Bigr)$ if $a > 49$. $\tag{$\ddagger$}$
Combining $(\dagger)$ and $(\ddagger)$, we can conclude that $49$ is the least integer value of $a$ for which algorithm $A'$ is asymptotically faster than algorithm $A$.
Best Answer
Since this question is asked frequently I will try to work out a solution for generic positive integers $a$ where $a\ge 2$.
Suppose we have $T(0)=0$ and $$T(1)=T(2)=\ldots =T(a-1)=1$$ and for $n\ge a$ $$T(n) = a T(\lfloor n/a \rfloor) + n \lfloor \log_a n \rfloor.$$
Furthermore let the base $a$ representation of $n$ be $$n = \sum_{k=0}^{\lfloor \log_a n \rfloor} d_k a^k.$$
Then we can unroll the recurrence to obtain the following exact formula for $n\ge a$ $$T(n) = a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times \sum_{k=j}^{\lfloor \log_a n \rfloor} d_k a^{k-j}.$$
Now to get an upper bound consider a string consisting of the digit $a-1$ to obtain $$T(n) \le a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times \sum_{k=j}^{\lfloor \log_a n \rfloor} (a-1) \times a^{k-j}.$$
This simplifies to $$a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times (a-1) \sum_{k=0}^{\lfloor \log_a n \rfloor-j} a^k$$ which is $$a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) (a^{\lfloor \log_a n \rfloor + 1 -j} -1)$$ which turns into $$a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} (\lfloor \log_a n \rfloor - j) (a^{\lfloor \log_a n \rfloor + 1} - a^j).$$ The sum produces four terms.
The first is $$\lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor + 1}.$$ The second is $$- \lfloor \log_a n \rfloor \frac{a^{\lfloor \log_a n \rfloor}-1}{a-1}.$$ The third is $$- \frac{1}{2} a^{\lfloor \log_a n \rfloor + 1} (\lfloor \log_a n \rfloor -1) \lfloor \log_a n \rfloor$$ and the fourth is $$\frac{1}{(a-1)^2} \left(a + a^{\lfloor \log_a n \rfloor} (\lfloor \log_a n \rfloor (a-1) -a)\right).$$
This bound represented by these four terms plus the leading term is actually attained and cannot be improved upon. For the asymptotics we only need the dominant term, which is $$\left(a - \frac{1}{2} a \right) \lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor} = \frac{1}{2} a \lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor}.$$
Now for the lower bound, which occurs with a one digit followed by zeroes to give $$T(n) \ge a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} a^j \times (\lfloor \log_a n \rfloor - j) \times a^{\lfloor \log_a n \rfloor-j}.$$ This simplifies to $$a^{\lfloor \log_a n \rfloor} + \sum_{j=0}^{\lfloor \log_a n \rfloor -1} (\lfloor \log_a n \rfloor - j) \times a^{\lfloor \log_a n \rfloor}$$ which is $$a^{\lfloor \log_a n \rfloor} + a^{\lfloor \log_a n \rfloor} \sum_{j=0}^{\lfloor \log_a n \rfloor -1} (\lfloor \log_a n \rfloor - j)$$ which finally produces $$a^{\lfloor \log_a n \rfloor} + a^{\lfloor \log_a n \rfloor} \sum_{j=1}^{\lfloor \log_a n \rfloor} j$$ or $$a^{\lfloor \log_a n \rfloor} + \frac{1}{2} \lfloor \log_a n \rfloor (\lfloor \log_a n \rfloor +1) a^{\lfloor \log_a n \rfloor}.$$ The dominant term here is $$\frac{1}{2} \lfloor \log_a n \rfloor^2 a^{\lfloor \log_a n \rfloor}.$$
Joining the dominant terms of the upper and the lower bound we obtain the asymptotics $$\color{#00A}{\lfloor \log_a n \rfloor^2 \times a^{\lfloor \log_a n \rfloor} \in \Theta\left((\log_a n)^2 \times a^{\log_a n}\right) = \Theta\left((\log n)^2 \times n\right)}.$$
This MSE link has a series of similar calculations.