Suppose we start by solving the following recurrence:
$$T(n) = T(\lfloor n/2 \rfloor) + 2T(\lfloor n/4 \rfloor) + n$$
where $T(1) = 1$ and $T(0) = 0.$
Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
We unroll the recursion to obtain an exact formula for $n\ge 1$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-z-2z^2}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
Studying the generating function we use partial fractions by residues
on the simple poles to get
$$ \frac{1}{1-z-2z^2}
= \frac{1}{z+1}\frac{1}{-1-4(-1)}
+ \frac{1}{z-1/2}\frac{1}{-1-4(1/2)}
\\ = \frac{1}{3} \frac{1}{z+1}
-\frac{1}{3} \frac{1}{z-1/2}
= \frac{1}{3} \frac{1}{z+1}
+ \frac{2}{3} \frac{1}{1-2z}.$$
This yields
$$ [z^j] \frac{1}{1-z-2z^2}
= \frac{1}{3} (-1)^j + \frac{2}{3} 2^j.$$
The new exact formula becomes
$$T(n) = \frac{1}{3}\sum_{j=0}^{\lfloor \log_2 n \rfloor}
(-1)^j \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}
+ \frac{2}{3}\sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
We now compute lower and upper bounds which are actually attained and
cannot be improved upon. For the lower bound consider a one digit
followed by a string of zeroes, to give
$$T(n) \ge \frac{1}{3}\sum_{j=0}^{\lfloor \log_2 n \rfloor}
(-1)^j 2^{\lfloor \log_2 n \rfloor-j}
+ \frac{2}{3}\sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j 2^{\lfloor \log_2 n \rfloor-j}
\\ = \frac{1}{3} 2^{\lfloor \log_2 n \rfloor}
\frac{1-(-1/2)^{\lfloor \log_2 n \rfloor+1}}{1-(-1/2)}
+ \frac{2}{3} (\lfloor \log_2 n \rfloor+1)
2^{\lfloor \log_2 n \rfloor}
\\ = \frac{2}{9} \left(2^{\lfloor \log_2 n \rfloor}
+ \frac{1}{2} (-1)^{\lfloor \log_2 n \rfloor}\right)
+ \frac{2}{3} (\lfloor \log_2 n \rfloor+1)
2^{\lfloor \log_2 n \rfloor}.$$
We see that the dominant term here is
$$\frac{2}{3} \lfloor \log_2 n \rfloor
2^{\lfloor \log_2 n \rfloor}.$$
For an upper bound consider a string of one digits to get
$$T(n) \le
\frac{1}{3}\sum_{j=0}^{\lfloor \log_2 n \rfloor}
(-1)^j \left(2^{\lfloor \log_2 n \rfloor - j + 1} - 1\right)
+ \frac{2}{3}\sum_{j=0}^{\lfloor \log_2 n \rfloor}
2^j \left(2^{\lfloor \log_2 n \rfloor - j + 1} -1 \right).$$
Recognizing the terms we already computed in the lower bound this
becomes
$$T(n) \le
\frac{4}{9} \left(2^{\lfloor \log_2 n \rfloor}
+ \frac{1}{2} (-1)^{\lfloor \log_2 n \rfloor}\right)
+ \frac{4}{3} (\lfloor \log_2 n \rfloor+1)
2^{\lfloor \log_2 n \rfloor}
\\ - \frac{1}{6} \left(1+(-1)^{\lfloor \log_2 n \rfloor}\right)
- \frac{2}{3} \left(2^{\lfloor \log_2 n \rfloor+1}-1\right).$$
Collecting like terms finally yields
$$T(n) \le
\frac{4}{9} 2^{\lfloor \log_2 n \rfloor}
+ \frac{1}{18} (-1)^{\lfloor \log_2 n \rfloor}
+ \frac{1}{2} +
\frac{4}{3} \lfloor \log_2 n \rfloor
2^{\lfloor \log_2 n \rfloor}.$$
We see that the dominant term here is
$$\frac{4}{3} \lfloor \log_2 n \rfloor
2^{\lfloor \log_2 n \rfloor}.$$
Joining the upper and the lower bound we get for the asymptotics of
this recurrence that it is
$$T(n)\in\Theta
\left(\lfloor \log_2 n \rfloor 2^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(\log_2 n \times 2^{\ \log_2 n}\right)
= \Theta(n \log n),$$
which, let it be said, could also have been obtained by inspection or
from the Master theorem.
Remark. Note that on the lower bound the floor on the logarithm is
exact and we get
$$\frac{2}{3} \log_2 n
2^{\log_2 n} = \frac{2}{3} n \log_2 n.$$
On the upper bound the floor on the logarithm is off by about one and
we get
$$\frac{4}{3} (\log_2 n - 1)
2^{\log_2 n - 1}
= \frac{2}{3} n \log_2 n - \frac{2}{3} n$$
which permits us to say that the dominant asymptotic is
$$\frac{2}{3} n \log_2 n.$$
This formula is very stable numerically which isn't always the case
with approximations to these Master theorem type recurrences.
A closely related recurrence where the asymptotics are of a different type is at this MSE link.
