Given $\alpha\in\Bbb N$, can there be more than $(\log N)^4$ divisors (composites allowed) of $N$ in $\big[\frac\alpha2,\alpha\big]$ when $\sqrt N\in\big[\frac\alpha2,\alpha\big]$?
What is the maximum number of divisors (composites allowed) asymptotically?
This is certainly possible. I will construct an example with $\alpha=\sqrt{N}$. That is, I will exhibit a square number $N$ with more than $(\log N)^4$ divisors lying in $[\tfrac{1}{2}\sqrt{N},\sqrt{N}]$.
Let $x>2$ be a large parameter. Consider $$ M:=\prod_{p\leq x}p \qquad\text{and}\qquad N:=M^2\lfloor\tfrac{2}{3}\log M\rfloor^{10}, $$ where the product is over the primes up to $x$. Note that $x\sim\log M$ by the prime number theorem. Hence, for large $x$, $M$ is divisible by any product $p_1p_2p_3p_4p_5$, where $p_i$ are distinct primes with $$ \tfrac{3}{5}\log M < p_1 < p_2 < p_3 < p_4 < p_5 < \tfrac{2}{3}\log M. $$ The number of such divisors $p_1p_2p_3p_4p_5$ of $M$ is $\gg(\log M/\log\log M)^5$ by the prime number theorem, i.e. more than $(\log N)^4$ when $x$ is large. Each of these divisors uniquely determines the divisor $Mp_1p_2p_3p_4p_5$ of $N$, and this divisor also satisfies $$ \tfrac{1}{2}\sqrt{N} < M\lceil\tfrac{3}{5}\log M\rceil^5 < Mp_1p_2p_3p_4p_5 < M\lfloor\tfrac{2}{3}\log M\rfloor^{5} =\sqrt{N}.$$ Done.
Added. Here I address the OP's second question which was added after my response above.
I claim that, for any $c<(\log 2)/3$, there are infinitely many square numbers $N$ with more than $\exp(c\log N/\log\log N)$ divisors lying in $[\tfrac{1}{2}\sqrt{N},\sqrt{N}]$. This is best possible up to the constant, because $c>\log 2$ would contradict Wigert's classical bound for the total number of divisors $\tau(N)$.
For the proof we present a variant of the construction above. We take $x$ and $M$ as before, and we decompose the interval $[1,\lceil\sqrt{M}\rceil]$ into $\ll\log M$ intervals of the form $[K,L]$ with integers $1\leq K < L \leq 2K$. Clearly one of these intervals, say $[K,L]$, contains $\gg\tau(M)/\log M$ divisors $d$ of $M$. Now we put $N:=M^2 L^2$, and we observe that each divisor $d$ considered above uniquely determines the divisor $Md$ of $N$. Moreover, this modified divisor $Md$ satisfies $$ \sqrt{N}/2=ML/2\leq MK\leq Md\leq ML=\sqrt{N}, $$ hence to finish the proof it suffices to remark that, for large $x$, $$ \tau(M)/\log M = \exp((\log 2+o(1))\log M/\log\log M)>\exp(c\log N/\log\log N).$$ This inequality follows from the prime number theorem combined with the bounds $N<2M^3$ and $3c<\log 2$. Done again.
