In Quirk and in Cirq, the convention used is that the +1 eigenvalue square roots into 1 (as opposed to -1) and that the -1 eigenvalue square roots into $i$ (as opposed to $-i$). More generally, you pick some convention for mapping a unitary eigenvalue $c$ into an angle $\theta$ where $e^{i \theta} = c$, and then define the gate raised to the power $p$ to use the corresponding eigenvalue $e^{i p \theta}$.
For example, the Z operation decomposes like this:
$Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = (+1) |0\rangle \langle 0| + (-1) |1\rangle \langle 1|$
Where the +1 and the -1 on the right hand side are the eigenvalues.
Square rooting the operation will square root the eigenvalues. +1 says +1 and -1 becomes $i$:
$\sqrt{Z} = \sqrt{+1} |0\rangle \langle 0| + \sqrt{-1} |1\rangle \langle 1| = (+1) |0\rangle \langle 0| + (i) |1\rangle \langle 1| = \begin{bmatrix} 1 & 0 \\ 0 & i\end{bmatrix}$
Now let's do $X$:
$X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = (+1) |+\rangle \langle +| + (-1) |-\rangle \langle -|$
where $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$
So:
$\sqrt{X} = (+1) |+\rangle \langle +| + (i) |-\rangle \langle -| = \begin{bmatrix} 0.5 + 0.5i & 0.5 - 0.5i \\ 0.5 - 0.5i & 0.5 + 0.5i\end{bmatrix}$
And now $Y$:
$Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = (+1) |i\rangle \langle i| + (-1) |-i\rangle \langle -i|$
where $|i\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ and $|-i\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$
So:
$\sqrt{Y} = (+1) |i\rangle \langle i| + (i) |-i\rangle \langle -i| = \begin{bmatrix} 0.5 + 0.5i & -0.5 - 0.5i \\ 0.5 + 0.5i & 0.5 + 0.5i\end{bmatrix}$
Another way to end up at the same convention is to start from the convention that $\sqrt{Z} = S = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}$, extend to other exponents using continuity and $Z^{a+b} = Z^{a} Z^{b}$, extend to $X$ by declaring that $X = HZH$ generalizes into $X^t = H Z^t H$, and extend to $Y$ using the right hand rule. Alternatively, you can extend to $Y$ by defining the "swap Y for Z" operation $H_{yz} = (Y + Z) / \sqrt{2}$ and declaring $Y^t = H_{yz} Z^t H_{yz}$.