Suppose we start by solving the following recurrence:
$$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + 4n$$
where $T(1) = c$ 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 $n\ge 2$
$$T(n) = c [z^{\lfloor \log_2 n \rfloor}] \frac{1}{1-z-z^2}
+ 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
[z^j] \frac{1}{1-z-z^2}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
We recognize the generating function of the Fibonacci numbers, so the
formula becomes
$$T(n) = c F_{\lfloor \log_2 n \rfloor +1}
+ 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
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
$$T(n) \ge c F_{\lfloor \log_2 n \rfloor +1}
+ 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{\lfloor \log_2 n \rfloor-j}
\\ = c F_{\lfloor \log_2 n \rfloor +1}
+ 8 \times 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{-j-1}.$$
Now since $$|\varphi|=\left|\frac{1+\sqrt{5}}{2}\right|<2$$
the sum term converges to a number, we have
$$\frac{1}{2} \le
\sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{-j-1}
\lt \sum_{j=0}^{\infty} F_{j+1} 2^{-j-1}
= 2.$$
For an upper bound consider a string of one digits to get
$$T(n) \le c F_{\lfloor \log_2 n \rfloor +1}
+ 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}
\\ = c F_{\lfloor \log_2 n \rfloor +1}
+ 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} (2^{\lfloor \log_2 n \rfloor+1-j} - 1)
\\ = c F_{\lfloor \log_2 n \rfloor +1}
- 4 (F_{\lfloor \log_2 n \rfloor +2} -1)
+ 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{\lfloor \log_2 n \rfloor+1-j}
\\ = c F_{\lfloor \log_2 n \rfloor +1}
- 4 (F_{\lfloor \log_2 n \rfloor +2} -1)
+ 4 \times 2^{\lfloor \log_2 n \rfloor+1}
\sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{-j}
\\ = c F_{\lfloor \log_2 n \rfloor +1}
- 4 (F_{\lfloor \log_2 n \rfloor +2} -1)
+ 16 \times 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{-j-1}.$$
The same constant appears as in the lower bound. Now since the term
$F_{\lfloor \log_2 n \rfloor}$ is asymptotically dominated by
$2^{\lfloor \log_2 n \rfloor}$ (we have $F_{\lfloor \log_2 n
\rfloor}\in o(2^{\lfloor \log_2 n \rfloor})$ because $F_{\lfloor \log_2 n
\rfloor} \in\Theta(\varphi^{\lfloor \log_2 n \rfloor}))$ joining the
upper and the lower bound we get for the asymptotics of this
recurrence that it is
$$T(n)\in\Theta\left(2^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(2^{\ \log_2 n}\right) = \Theta(n),$$
which, let it be said, could also have been obtained by inspection.
Remark. The evaluation of the constant is done by noting that
the generating function of
$$F_{j+1} 2^{-j-1}\quad
\text{is}\quad\frac{1/2}{1-z/2-z^2/4}$$
which at $z=1$ evaluates to $\frac{1/2}{1-1/2-1/4} = 2.$
We have a certain flexibility as to what power of two to use in the constant but this does not affect the asymptotics.
This MSE link has a similar calculation.
