Bounding the recurrence $f(n)=2f(n-1)+2f(n/2)$
For concreteness, let's assume that your recurrence is really
$$ f(n) = 2f(n-1) + f(\lfloor n/2 \rfloor) + f(\lceil n/2 \rceil). $$
Define $g(n) = f(n)/2^n$. Then
$$ g(n) = g(n-1) + 2^{\lfloor n/2 \rfloor + 1 - n} g(\lfloor n/2 \rfloor) + 2^{\lceil n/2 \rceil + 1 - n} g(\lceil n/2 \rceil). $$
In particular,
$$ g(n) \leq (1 + 2^{\lfloor n/2 \rfloor + 1 - n} + 2^{\lceil n/2 \rceil + 1 - n}) g(n-1). $$
Taking logarithms,
$$ \log g(n) \leq \log g(n-1) + O(2^{\lfloor n/2 \rfloor + 1 - n} + 2^{\lceil n/2 \rceil + 1 - n}). $$
We conclude that
$$ \log g(n) \leq O(1) + O(1) \sum_{m=c}^n [2^{\lfloor m/2 \rfloor + 1 - m} + 2^{\lceil m/2 \rceil + 1 - m}] = O(1), $$
where $c$ is some arbitrary starting point.
Therefore $\log g(n)$, and so $g(n)$, is bounded. Since clearly $g(n)$ is monotone increasing, we deduce that $g(n)$ tends to some constant $C$. We conclude that
$$ f(n) \sim C 2^n. $$