Let $a=t(1)$. Then
$$\begin{align*}
t(2^1)&=2^{2^1}a+2^1\;;\\
t(2^2)&=2^{2^2}\left(2^{2^1}a+2^1\right)+2^2\\
&=2^{2^2+2^1}a+2^{2^2+1}+2^2\;;\\
t(2^3)&=2^{2^3}\left(2^{2^2+2^1}a+2^{2^2+1}+2^2\right)+2^3\\
&=2^{2^3+2^2+2^1}a+2^{2^3+2^2+1}+2^{2^3+2}+2^3\;;\\
t(2^4)&=2^{2^4}\left(2^{2^3+2^2+2^1}a+2^{2^3+2^2+1}+2^{2^3+2}+2^3\right)+2^4\\
&=2^{2^4+2^3+2^2+2^1}a+2^{2^4+2^3+2^2+1}+2^{2^4+2^3+2}+2^{2^4+3}+2^4\;;\\
t(2^5)&=2^{2^5+2^4+2^4+2^3+2^2+2^1}a+2^{2^5+2^4+2^3+2^2+1}+2^{2^5+2^4+2^3+2}+2^{2^5+2^4+3}+2^{2^5+4}+2^5\;;\\
&\vdots\\
t(2^k)&=2^{2^{k+1}-2}a+\sum_{i=1}^k 2^{i+\sum_{j=i+1}^k 2^j}\\
&=2^{2^{k+1}-2}a+\sum_{i=1}^k 2^{i+2^{k+1}-2^{i+1}}\;,
\end{align*}$$
so if $n=2^k$,
$$\begin{align*}
t(n)&=2^{2n-2}a+\sum_{i=1}^k 2^{2n+i-2^{i+1}}\\
&=2^{2n-2}a+2^{2n}\sum_{i=1}^k 2^{i-2^{i+1}}\\
&=2^{2n-2}a+2^{2n}\frac{2^{k+1}-2}{4\sum_{i=0}^{k-1}2^{2^i}}\\
&=2^{2n-2}\left(a+\frac{2n-2}{\sum_{i=0}^{k-1}2^{2^i}}\right)\;.
\end{align*}$$
Now $$2^{n/2}=2^{2^{k-1}}\le\sum_{i=0}^{k-1}2^{2^i}<2^{2^k}=2^n\;,$$ so
$$2^{2n-2}\left(a+\frac{2n-2}{2^n}\right)
or $$2^{2n-2}a+(2n-2)2^{n-2}