How to show $\frac{d}{d x}\left(|x|^{1/2}\right)=\frac{x}{2|x|^{3/2}}$?
Let $f(x) = |x|^{1/2}$. Of course, away from $x=0$, the answer of Robert Lee (without any Dirac delta) is correct, and you will just have $$ \begin{align*} f'(x) &= \frac{x}{2\,|x|^{3/2}} \\ f''(x) &= \frac{-1}{4\,|x|^{3/2}} \end{align*} $$ for any $x\in\mathbb{R}\setminus\{0\}$ (since $x^2/|x|^{7/2} = 1/|x|^{3/2}$).
So, to know if there is a Dirac delta, you have to take the derivative in the sense of distributions. To get the first derivative, since $f'(x) = \frac{x}{2\,|x|^{3/2}}$ everywhere except at $x=0$, we see that $f'(x) = \frac{x}{2\,|x|^{3/2}}$ almost everywhere. Moreover, since $|f'| = \frac{1}{2\,|x|^{1/2}}$ is locally integrable, we obtain that the identity $f'(x) = \frac{x}{2\,|x|^{3/2}}$ is also valid locally in $L^1$ and so also in the sense of distributions.
This means that to compute $f''$ in the sense of distributions, we just need to compute the derivative of $f'$ in the sense of distributions. Let $\varphi$ be a smooth compactly supported (let say $\varphi = 0$ in $[-a,a]$ for some $a>0$) test function. Then by definition and the above computation $$ \langle f'',\varphi\rangle = \langle(f')',\varphi\rangle = -\langle f',\varphi'\rangle = -\langle \tfrac{x}{2\,|x|^{3/2}},\varphi'\rangle $$ and we can write this as an integral since both functions are locally integrable. Therefore $$ \begin{align*} \langle f'',\varphi\rangle &= -\int_{\mathbb{R}} \tfrac{x}{2\,|x|^{3/2}}\,\varphi'(x)\,\mathrm{d}x = -\int_{-a}^a \tfrac{x}{2\,|x|^{3/2}}\,\varphi'(x)\,\mathrm{d}x \\ &= -\int_{-a}^a \tfrac{x}{2\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))'\,\mathrm{d}x \end{align*} $$ Since $\varphi(x)-\varphi(0)$ is vanishing at $x=0$ and even better, $|\varphi(x)-\varphi(0)| ≤ C\,|x|$ (by the fundamental theorem of calculus, with $C= \sup |\varphi'|$ for instance), we can integrate by parts and use the value of $f''$ that was already computed for $x≠0$ to get $$ \begin{align*} \langle f'',\varphi\rangle &= \left[-\tfrac{x}{2\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\right]_{-a}^a - \int_{-a}^a \tfrac{1}{4\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x \\ &= - \int_{-a}^a \tfrac{1}{4\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x \end{align*} $$ since $\varphi(a) = \varphi(-a) = 0$. The distribution $\mathrm{pf}(\frac{1}{|x|^a})$ for $a\in(1,2)$ is known as the finite part of $\frac{1}{|x|^a}$ and is defined exactly by $$ \begin{align*} \langle \mathrm{pf}(\tfrac{1}{|x|^a}),\varphi\rangle &= \int_{\mathbb{R}} \tfrac{1}{|x|^{a}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x \end{align*} $$ so in the sense of distributions $$ f'' = \frac{-1}{4} \mathrm{pf}\left(\frac{1}{|x|^{3/2}}\right) $$
Remarks: I do not know where you found the formula with a Dirac delta but it is obviously wrong, since there is no meaning to $\frac{\delta(x)}{\sqrt{x}}$.
Even if it is true that $\mathrm{sign}' = \delta_0$ in the sense of distributions, the following computation $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x}{|x|^{3/2}}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{sign}(x)\frac{1}{|x|^{1/2}}\right) = 2\delta_0(x)\left(\frac{1}{|x|^{1/2}}\right) + \mathrm{sign}(x)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{|x|^{1/2}}\right) $$ is not valid since one cannot multiply singular distributions in general.
This is not a complete answer, but for the case of the absolute function defined on the real numbers, this might help.
Since $|x|$ is non-differentiable only at $0$, all the derivatives we calculate will have domain $D \subseteq\mathbb{R} \setminus \{0\}$.
If we write $|x|=\sqrt{x^2}$ we see that $$ \frac{d}{dx} \left(|x| ^\frac{1}{2} \right) = \frac{d}{dx} \left(x^2\right)^\frac{1}{4} = \frac{1}{4}\left(x^2\right)^{-\frac{3}{4}} \left[\frac{d}{dx}x^2 \right]= \frac{1}{4}\left(x^2\right)^{-\frac{3}{4}}2x = \frac{x}{2\left(x^2\right)^\frac{3}{4}} = \frac{x}{2\left(\sqrt{x^2}\right)^\frac{3}{2}} = \frac{x}{2|x|^\frac{3}{2}} $$ As for the second part $$ \frac{d^2}{dx^2} \left(|x| ^\frac{1}{2} \right) = \frac{d}{dx} \left(\frac{x}{2\left(x^2\right)^\frac{3}{4}}\right) = \frac{2\left(x^2\right)^\frac{3}{4} - 2x\left(\frac{3}{4}(x^2)^{-\frac{1}{4}} (2x)\right)}{4\left(x^2\right)^\frac{3}{2}} = \frac{2\left(x^2\right)^\frac{3}{4}}{4\left[\left(x^2\right)^\frac{3}{4}\right]^2} - \left(\frac{3}{4}\right)\frac{4x^2}{4\left(x^2\right)^\frac{3}{2}(x^2)^{\frac{1}{4}}} = \frac{1}{2\left(x^2\right)^\frac{3}{4}}\left(\frac{2x^2}{2x^2}\right) - \frac{3x^2}{4\left(x^2\right)^\frac{7}{4}} = \frac{2x^2 -3x^2} {4\left(x^2\right)^\frac{7}{4}} = -\frac{x^2}{4 |x|^\frac{7}{2}} $$ Now, since the Dirac delta turns it's input into $0$ if $x \neq 0$, and the domain on which we're working on doesn't have $0$, our last answer will be equal to $$ -\frac{x^2}{4 |x|^\frac{7}{2}} = 0 -\frac{x^2}{4 |x|^\frac{7}{2}} = \frac{\delta(x)}{\sqrt{|x|}} -\frac{x^2}{4 |x|^\frac{7}{2}} $$ since $\frac{\delta(x)}{\sqrt{|x|}}=0$ for all $x \in D$.
I suspect there might be a more general definition of the absolute value you can work with to get the Dirac delta out directly, but I'm not familiar with it. Hope this helps!
$$\sqrt{\pm x}\to\pm\frac1{2\sqrt{\pm x}}$$ so
$$\sqrt{|x|}\to\text{sgn}(x)\frac1{2\sqrt{|x|}}.$$
For the second derivative, you can use the same method. But as there is a discontinuity, you can represent its derivative with a $\delta$.