I am trying to understand how $$ \mathbb{P}\left(-z_{\alpha / 2} \leqslant \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \leqslant z_{\alpha / 2}\right) \approx 1-\alpha $$
can be converted to $$ \mathbb{P}\left(\bar{X}-\frac{z_{\alpha / 2}}{\sqrt{n}} \cdot \sigma \leqslant \mu \leqslant \bar{X}+\frac{z_{\alpha / 2}}{\sqrt{n}} \cdot \sigma\right) \approx 1-\alpha $$
In order to make $\mu$ positive I tried multiplying everything by $-1$. But I understand that causes the greater than signs to switch.
I'll assume that $X \sim N(\mu, \sigma^2)$, then, $\frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$, therefore,
$$P( Z_{\alpha / 2} \leq \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \leq Z_{1 - \alpha/2}) = 1-\alpha $$
Some algebra,
$$P( \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} \leq \bar{X} - \mu \leq \frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2}) = 1-\alpha $$
Subtracting the average,
$$P( \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} - \bar{X} \leq - \mu \leq \frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} - \bar{X}) = 1-\alpha $$
Now we multiply by -1, so we need to change the direction of the inequalities,
$$P( - \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} + \bar{X} \geq \mu \geq -\frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} + \bar{X}) = 1-\alpha $$
Rearrange,
$$P( \bar{X} -\frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} \leq \mu \leq \bar{X} - \frac{\sigma}{\sqrt{n}} Z_{\alpha / 2} ) = 1-\alpha $$
We now note, that $Z_{\alpha /2} = -Z_{1- \alpha/ } $ due to the symmetry of the standard normal distribution with respect to 0
$$P( \bar{X} -\frac{\sigma}{\sqrt{n}}Z_{1 - \alpha/2} \leq \mu \leq \bar{X} + \frac{\sigma}{\sqrt{n}} Z_{1 - \alpha / 2} ) = 1-\alpha, $$
which means the CI can also be written as $\bar{X} \pm \frac{\sigma}{\sqrt{n}} Z_{1 - \alpha / 2} $.
You're right about the signs switching. The symmetry of the normal distribution may be confusing you.
If $X \sim \mathsf{Norm}(\mu, \sigma),$ then $Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \mathsf{Norm}(0,1).$
In order to make a 95% CI, you can find numbers $L$ and $U$ such that $P(L < Z < U) = 0.95.$
Then the inequality inside the probability statement becomes $$L < \frac{\bar X - \mu}{\sigma/\sqrt{n}} < U$$ or, upon "pivoting," $$\bar X - U\frac{\sigma}{\sqrt{n}} < \mu < \bar X - L\frac{\sigma}{\sqrt{n}}.$$
It is customary to cut the same probability $0.025$ from each tail of the symmetrical normal distribution, so $L = -1.96, U = 1.96$ and plugging those values into the second displayed inequality above becomes
$$\bar X - 1.96\frac{\sigma}{\sqrt{n}} < \mu < \bar X + 1.96\frac{\sigma}{\sqrt{n}}.$$
