Let me write $a\approx b$ for $a=\Theta(b)$, and $t=\frac n2(1+\tau)$.

Stirling approximation gives
$$2^{-n}\binom nt=(1+o(1))\frac{n^n}{(2t)^t(2(n-t))^{n-t}}\sqrt{\frac n{2\pi t(n-t)}}
\approx\bigl((1+\tau)^{1+\tau}(1-\tau)^{1-\tau}\bigr)^{-n/2}\frac1{\sqrt{n-t}}.$$
For any $t\le k<t+l$, we have
$$\binom n{k+1}\binom nk^{-1}=\frac{n-k}{k+1}\le\frac{n-t}{t+1}\le\frac{1-\tau}{1+\tau},$$
hence
$$\binom nt^{-1}\sum_{k=t}^{t+l}\binom nk\le\sum_{k=0}^l\left(\frac{1-\tau}{1+\tau}\right)^k=\frac{1+\tau}{2\tau}\left(1-\left(\frac{1-\tau}{1+\tau}\right)^{l+1}\right)\approx\frac1\tau\left(1-\left(\frac{1-\tau}{1+\tau}\right)^{l+1}\right).$$
Similarly, if we put $\tau'=\tau+2l/n$ so that $t+l=\frac n2(1+\tau')$, we have
$$\binom n{k+1}\binom nk^{-1}\ge\frac{1-\tau'}{1+\tau'},$$
hence
$$\binom nt^{-1}\sum_{k=t}^{t+l}\binom nk\ge\sum_{k=0}^l\left(\frac{1-\tau'}{1+\tau'}\right)^k\approx\frac1\tau\left(1-\left(\frac{1-\tau'}{1+\tau'}\right)^{l+1}\right),$$
where I’ve used the fact that $\tau\le\tau'\le2\tau$ due to the assumptions.

If $l\ge1/\tau\ge1/\tau'$, then
$$\left(\frac{1-\tau}{1+\tau}\right)^{l+1}\le(1+o(1))e^{-2}<1,$$
and likewise for $\tau'$, hence both bounds reduce to
$$\binom nt^{-1}\sum_{k=t}^{t+l}\binom nk\approx\frac1\tau.$$
On the other hand, if $l\le1/\tau\le2/\tau'$, then
$$\left(\frac{1-\tau'}{1+\tau'}\right)^k\ge(1+o(1))e^{-4},\qquad k\le l,$$
hence
$$\binom nt^{-1}\sum_{k=t}^{t+l}\binom nk\approx l.$$
Putting everything together, and using the fact that $n-t\approx n$ by assumption, we obtain
$$2^{-n}\sum_{k=t}^{t+l}\binom nk\approx\frac{\min\{l,\tau^{-1}\}}{\sqrt n}\bigl((1+\tau)^{1+\tau}(1-\tau)^{1-\tau}\bigr)^{-n/2}.$$