Given a uniform prior and (independent) observations from a Normal distribution then the resulting posterior is a truncated normal distribution. However, in this case the observations are drawn from a truncated prior which makes it more complicated.
First, you can 'ignore' the integral in the denominator since this is just a constant assuring that the posterior is a density. In general
$$p(\mu | x) \propto p(x|\mu)p(\mu).$$
As you have derived (note that $1/\sigma$ is a constant and is not considered):
$$p(\mu|x) \propto \frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\Phi\left(\frac{1-\mu}{\sigma}\right) - \Phi\left(\frac{-\mu}{\sigma}\right)}I_{\mu \in [0,1]}.$$
At first glance it looks like it is a truncated normal again, however $\mu$ is now variable instead of $x$, so comparing with the truncated normal density, this is no longer the case.
Your interpretation of $Z = \frac{X - \mu}{\sigma}$ is correct. For example,
IQ scores are sometimes considered as being distributed $\mathsf{Norm}(\mu=100,\sigma=15).$ Then if someone gets a 'raw' test score of $x = 115,$ one
can say that their 'standard' score is $z = 1,$ meaning that they scored
one standard deviation above the mean.
Then if we want to know the probability of a raw score less than $x \le 115,$ one can use the standard score to get that probability:
$$P(X \le 115) = P\left(\frac{X = \mu}{\sigma} \le \frac{115 - 100}{15}\right)
= P(Z \le 1) = 0.8413,$$
where you can get the probability 0.8413 from a printed table of the normal
distribution. If your table is a pure CDF table then you can read 0.8413 directly from the table. If your table shows the probability between 0 and z, you will
see 0.3413 in the table, to which you'd need to add 0.5 to get 0.8413.
In symbols,
$$P(Z \le 1) = P(Z \le 0) + P(0 < Z \le 1) = 0.5000 + 0.3413 = 0.8413.$$
We have $P(Z \le 0) = 0.5000$ because the standard normal density curve is
symmetric about $0$ so half of the area under the curve (representing probability) lies on either side of $0.$
In the figure below, the density function of $\mathsf{Norm}(100,15)$ is shown
on the left and the standard normal density curve is shown on the right.
In both curves the area between the vertical red lines represents the
probability 0.3413.
Sometimes one says that the raw score 115 lies at about the 84th percentile
because about 84% of the population gets raw scores below 115 on an IQ test.
It is possible to use a table for the standard normal random variable $Z$ to
get probabilities relating to the random variable $X \sim \mathsf{Norm}(100,15)$ because all normal distributions have the same "fundamental shape."
When you speak in terms of standard deviations above and below the mean,
that applies to any normal distribution. Just as $P(Z \le 1) = 0.8413$
and $P(X \le 115) = 0.8413,$ it is also true that $Y \sim \mathsf(\mu = 50, \sigma = 2)$ has $P(Y \le 52) = 0.8413$ because the raw score $y = 52$ also
corresponds to standard score $z = 1.$
Using a printed table, it is necessary to use standard scores because
it would be impossible to print separate tables, one for each conceivable
value of $\mu$ and each conceivable value of $\sigma.$
Finally, it is not possible to use the standard normal density function
directly to get probabilities. You need a standard normal cumulative distribution (CDF) table for that. For some distributions, you can use
the density function to get probabilities by integration, but that
does not work for the normal distribution. (Informally, you might say that the
expression $\varphi(z) = \frac{1}{\sqrt{2\pi}}e^{-0.5x^2}$ is "too messy"
to integrate, using the usual methods of calculus.)
This means that it is tedious to make a normal table and you should be glad someone has done it for you. Roughly speaking,
one has to break the desired region into rectangles and sum the areas of
the rectangles. One could get 'pretty close" to 0.3413 by summing the
areas of the five rectangles between the vertical red bars in the figure below.
The height of each rectangle (at its center) can be found from the standard normal density
function $\varphi(z)$ and the widths are all 0.02.
Best Answer
Let $\sigma ^ 2 = \theta $, thus $ X \sim N( \mu, \theta)$, hence $$ f_X(x; \theta) = \frac{1}{\sqrt{ 2 \pi \theta }} \exp\left( \frac {- (x - \mu ) ^ 2} { 2\theta} \right), $$
$$ l(\theta) = - \tfrac 1 2 \ln \theta - \frac {(x - \mu )^2} {2\theta} + \text {constant} $$ $$ l'(\theta) = -\frac{1}{2\theta} + \frac{(x- \mu) ^2}{2\theta ^ 2} $$ $$ - \mathbb{E} l'' (\theta) = - \mathbb{E}[ \frac{1}{2\theta ^ 2} - \frac{(x- \mu) ^2}{\theta ^ 3} ] = -\frac{1}{2\theta ^ 2} + \frac{1}{\theta^2} = \frac{1}{2 \theta ^ 2}. $$ Use the additive property of Fisher's information to get the Info. for sample of size $n$, i.e., $$ I_{X_1,...,X_n}(\theta) = \frac{n}{2\theta ^ 2} = \frac{n}{2\sigma ^ 4}, $$ for the observed information replace $\sigma ^2 $ with $$ S ^ 2 = \frac{\sum_{i=1}^n ( X_i - \mu) ^ 2}{n}. $$ (And note that $\operatorname{var}(X) = \mathbb{E}(X - \mu ) ^2 = \sigma ^ 2$).