What does it mean that a random variable is uniformly distributed between two practically speaking? Say I have the explicit density function of $X$, and I'm told that $T$ is a random variable that is uniformly distributed between, say $4X$ and $6X$, or $1$ and $X^2+2$, how can I derive the density function, the expected value of $T$, based on this information?

This is what I'm sure it means: If $T$ is uniformly distributed between $X$ and $Y$ then $$Exp(T|X)=\frac{1}{2}Exp(X+Y)$$. For the density function I don't know how to derive it at all, but I suspect there might be a trick to first find the cumulative distribution and then differentiate.
 A: For a concrete example, let's say $X\sim U(0,1)$ and "$T\sim U(0,X)$". Using random variables as parameters when specifying the distribution of another random variable isn't really standard (as far as I know), but we can reasonably interpret this as describing a process where we first sample $x$ from a $U(0,1)$ distribution and then "create" a $U(0,x)$ distribution and sample from it.
More formally, this means we have a pair of random variables $X$ and $T$ with the property that the conditional distribution of $T$, given that $X=x$, is the $U(0,x)$ distribution. That is:
$$
f_{T|X}(t|x)=\begin{cases}\frac1x&\text{if $0<t<x$}\\0&\text{otherwise}\end{cases}
$$
so the unconditional distribution of $T$ is
$$
\begin{align}
f_T(t)
&=\int_0^1f_{X,T}(x,t)dx\\
&=\int_0^1f_X(x)f_{T|X}(t|x)dx\\
&=\int_t^1\frac1xdx\\
&=-\log t
\end{align}
$$
for $0<t<1$.
We can check that this is indeed a density function: $-\log t$ is positive on $(0,1)$, and $\int_0^1(-\log t)dt=1$. We also have $\Bbb ET=\int_0^1t(-\log t)dt=\frac14=\frac12(\Bbb E0+\Bbb EX)$, which matches your expectation. (No pun intended!)

Edit: As pointed out in a comment here, instead of defining $T$ by its conditional distribution, we can think of $T$ as the result of translating and scaling a new independent $U(0,1)$ random variable onto the desired interval. That is, $T\sim U(X,Y)$ means $T=X+(Y-X)Z$, where $Z\sim U(0,1)$ is independent from $X$ and $Y$. From this it follows immediately that $\Bbb ET=\frac12(\Bbb EX+\Bbb EY)$, but I think determining other properties of the distribution will involve the same calculus as above.
