$$s=ut+\frac{1}{2}at^2$$ Assuming, $u=0$
$$s=\frac{1}{2}at^2$$ $$a=\frac{2s}{t^2}$$
So I got a function for acceleration.
$$v=u+at$$ $$v=\frac{2s}{t} \tag{1}$$
In reality I know that $s=vt$ when velocity is constant. Here the tag 1 equation don't have anything to do with acceleration according to that equation. I believe both term will give different solution. So there's something wrong happening with these equation,isn't it? (Side-note : I know calculus but I was defining them in general term)
The relationships $s = s_{initial} + ut + {1 \over 2} a t^2$ and $v = u + at$ assume constant acceleration. (You left out the initial position $s_{initial}$ in your relationship.) Your results $a = {2s \over t^2}$ and $v = {2s \over t}$ are correct, assuming the initial position $s_{initial}$ and initial velocity $u$ are both zero.
$a$ is constant. But $v$ is not constant (unless $a$ is zero) since given acceleration the velocity $v(t) = at$ where $a$ is constant but t changes. So $s \ne vt$; $s = \int_{o}^{t} v(t) \enspace dt =\int_{o}^{t} at \enspace dt$, and with $a$ constant $s = {a t^2 \over 2}$.
If $a$ is zero, $v = u$, a constant, and $s = s_{initial} + ut$. If the initial position $s_{initial}$ is zero, $s = ut$ where $u$ is the constant initial velocity; that is $s = \int_{o}^{t} u \enspace dt$ where $u$ is constant.
Your expression is correct, as you can check by, e.g., plotting the expressions.
You explicitly assumed the initial velocity to be zero. If the velocity is constant, then the final formula will give the correct expression: the particle doesn't move and the position is constant.
If $v$ is constant then $a=0$. Further if you assume the initial velocity $u=0$, then $v=0$ (since there is no acceleration) and indeed $s=0$ since there is no displacement at a constant velocity of $0$. Thus the equations $v=2s/t$ and $v=s/t$ are perfectly consistent since $s=0$ on the right of each, and $v=0$ on the left of each.
