Is the parameter of interest uniformly distributed over the confidence interval? Let's assume that we are estimating a parameter of the population (e.g., mean height or weight). We've gathered our sample and calculated confidence intervals around the parameter of interest. Are we, therefore, assuming that this parameter follows uniform distribution over the confidence interval? In simple terms, is the parameter equally probable to be anywhere within this confidence interval?

Because you are asking about confidence intervals,
I will assume these are frequentist interval estimates.
A 90% t CI, based on the sample mean $\bar X$ and the sample
standard deviation $S$ is of the form
$$\bar X \pm t^*S/\sqrt{n},$$ where $t^*$ cuts probability $0.05$ from the upper tail of Student's t distribution with $n - 1$ degrees of freedom.
The analogous probability statement is
$$P\left(-t^* < \frac{\bar X - \mu}{S/\sqrt{n}} < t^*\right) = 0.9,$$
in which probability is not uniformly distributed in $(-t^*,t^*).$
Consequently, you cannot get a 45% CI just by making
that interval half as long: $(-0.5t^*, 0.5t^*).$
In order to get a 45% CI, you would need to cut probability
$0.275$ from each tail of $\mathsf{T}(\nu=99).$
Suppose you have the fictitious sample below, of size $n = 100$ from
a normal population with unknown mean $\mu$ and $\sigma,$
Then a 90% CI for $\mu$ would be $(59.79, 62.16)$
and a 45% CI would be $(60.54, 61.40),$ as computed using R below.
set.seed(2022)
x = rnorm(100, 60, 7)

CI.90= mean(x) + qt(c(.05,.95),99)*sd(x)/10; CI.90
[1] 59.78559 62.15685

CI.45= mean(x) + qt(c(.275,.725),99)*sd(x)/10; CI.45
[1] 60.54292 61.39953

An alternative way to get the 45% CI, would be to use the confidence
interval from the t.test procedure in R.
t.test(x, conf.lev=.45)$conf.int
[1] 60.54292 61.39953
attr(,"conf.level")
[1] 0.45

An incorrect interval $(60.38, 61.56)$ that simply makes CI.90half as
long, would not be a true 45% CI for $\mu.$
CI.u= mean(x) + .5*qt(c(.05,.95),99)*sd(x)/10; CI.u
[1] 60.37841 61.56404

