tl;dr: If $z_1 = g^t \pmod p$ is one of the square roots of $g^x \pmod p$ such that $ \frac{p-1}{2} \leq t < p-1$. Then, does $p-z_1 = g^{\frac{x}{2}} \pmod p$ hold true?
Say that we define a function and an input $y=g^x \pmod p$ (similar to the most significant bit of the exponent function):
$$ MSB(y) = \begin{cases} 0, & \text{if }{ 0<x <\frac{p-1}{2} } \\ 1, & \text{if }{ \frac{p-1}{2} \leq x < p-1 } \end{cases} $$
$y = g^x \pmod p$ and p is prime. If we have a square root finder procedure, SQRT(a), for quadratic residues (i.e. returns both solutions to $y = x^2 \pmod p$ ) and we obtain the square roots to $g^x$ (when $g^x$ is a Quadratic Residue), i.e:
$$ (z_1,z_2) \leftarrow SQRT(g^x) $$
Is it true that if $MSB(g^x) = 1$ then $ p - z_1 = g^{ \frac{x}{2}} \pmod p$.
Notice how taking square roots divides the exponent by 2 which corresponds bit shifting the exponent.
Context/Motivation for question (Cryptography and inverting discrete log):
The reason I am asking this is, if we had a way to compute efficiently the MSB function (some black box/oracle), then we could bit shift the exponent of a given element and extract the Least Significant Bit (LSB) of the exponent using the Legendre Symbol, because I would know if the power is even or odd. Thus, making the Discrete Log efficiently computable. Obviously big if on the part of computing the MSB.
Fix a primitive root $g$ and odd prime $p$. Since $g^{(p-1)/2}$ mod $p$ is a nontrivial square root of one it must be congruent to $-1$ mod $p$. Suppose $x,t\in\Bbb Z$ are such that $g^t$ is a square root of $g^x$. As $g^x$ is a square, $x$'s residue mod $p-1$ must be a multiple of two, so $x$ must be even, and $g^{x/2}$ is also a square root of $g^x$ mod $p$. Then $(g^{t-x/2})^2\equiv1$ so $g^{t-x/2}\equiv\pm1$ depending on if $t-x/2$ is congruent to either $0$ or $(p-1)/2$ mod $p-1$. This yields the conclusion
$$\begin{cases} g^{x/2}\equiv g^{t} & {\rm if}~\frac{t-x/2}{(p-1)/2}~{\rm is~even} \\ g^{x/2}\equiv -g^t & {\rm if}~\frac{t-x/2}{(p-1)/2}~{\rm is~odd}\end{cases} $$
Given $\frac{p-1}{2}\le t<p-1$ and $x\ge0$, for $\frac{t-x/2}{(p-1)/2}$ to be odd, it would have to be $1$, i.e. $t=\frac{(p-1)+x}{2}$. If you allow $x$ to be negative then more relationships between $t$ and $x$ are possible for this condition.