You need to pull $P(U>s)$ from the original distribution, not from the limited range of $(t,1)$. Therefore, $P(U>s) = \frac {1-s}{1-0}$. It's the length of the segment $[s,1]$ divided by the length of the whole segment. We then divide by the probability of the condition $P(U>a) = \frac {1-a}{1-0}$ and you'll have your final answer.
The key thing in conditional probability is that we pull the probabilities from the original distribution, not the new distribution based on the condition. We're actually calculating the new distribution based on the condition.
A concrete example using the same range:
What's the probability that U is larger than $\frac 34$ given that U is larger than $\frac 12$?
We can tell (relatively intuitively) that this will be equal to $\frac 12$. $\frac 34$ is halfway between $\frac 12$ and $1$. But to calculate directly using the formula:
$$P(U>\frac34|U>\frac12) = \frac {P(U>\frac34)}{P(U>\frac12)}$$
From the original range, $P(U>\frac34) = \frac14$ and $P(U>\frac12) = \frac12$. So:
$$P(U>\frac34|U>\frac12) = \frac{\frac14}{\frac12} = \frac12$$
Again, just make sure you're pulling the top probability from the original distribution.
The definition of a "uniform distribution" is that the density function is constant for all $x,y$ within the support region. So one must have
$$f_{X,Y}(x,y) = \frac{1}{A}$$
where $A$ is the area of either the square or the circle.
The same formula will hold for the density function of a "uniform distribution" on any geometric region.
Note though that this idea of uniform-ness applies only to the Cartesian coordinates ($x,y$). If you re-parametrized the circle in terms of radius and angle ($r,\theta$), for example, then the radial distance would not be uniformly distributed.
The angle $\theta$ would be uniformly distributed on $(0,2\pi)$ but the radial distance $r$ would have to follow a triangular distribution for the bivariate distribution to be uniform in terms of $x$ and $y$.
Best Answer
A picture might help. Independent uniform distributions on the interval $[0,1]$ may be considered a uniform distribution on the unit square $I^2 = [0,1]\times [0,1]$. Events are regions in the square and their probabilities are their areas.
Let $z$ be any possible value of $\max(U,V)$. The set of coordinates $(U,V)$ where $\max(U,V)=z$ forms the top and right edges of a square of side $z$. Let $dz$ be a small positive number. The set of coordinates $(U,V)$ whose maximum lies between $z$ and $z+dz$ forms a narrow thickening of that square, as shaded in the figure. Its area is the difference of the areas of two squares, one of side $z+dz$ and the other of side $z$, whence
$$\Pr(z \le Z \le z+dz) = (z+dz)^2 - z^2 = 2z\,dz + (dz)^2.\tag{1}$$
Let $u$ be any possible value of $U$: it is marked with a vertical dashed line in the figures.
The left panel shows a case where $u \le z$: The chance that $U\le u$ would be the area to the left of that line (equal to $u$); but the event that $U\le u$ and $Z$ lies between $z$ and $z+dz$ is just the brown shaded area. It's a rectangle, so its area is its width $u$ times its height $dz$. Thus,
$$\Pr(U \le u, z \le Z \le z+dz) = u\,dz.\tag{2}$$
The right panel shows a case where $z \lt u \le z+dz$. Now the chance that $U \le u$ and $z \lt Z \le z+dz$ consists of two rectangles. The top one has base $u$ and height $dz$; the right one has base $(u-z)$ and height $z+dz$. Therefore
$$\Pr(U \le u, z \le Z \le z+dz) = u\, dz + (u-z)(z+dz).\tag{3}$$
By definition, the conditional probabilities are these chances divided by the total chance that $z \le Z \le z+dz$, given in $(1)$ above. Divide $(2)$ and $(3)$ by this value. Letting $dz$ be infinitesimal, and retaining the standard part of the result, gives the chances conditional on $Z=z$. Thus, when $0 \le u \le z$,
$$\Pr(U \le u\,|\, Z=z) = \frac{u\,dz}{2z\,dz + (dz)^2} = \frac{u}{2z + dz} \approx \frac{u}{2z}.$$
When $z \lt u \le z+dz$, write $u = z + \lambda dz$ for $0 \lt \lambda \le 1$ and compute
$$\Pr(U \le u|Z=z) = \frac{u\, dz + (u-z)(z+dz)}{2z\,dz + (dz)^2} = \frac{(z + \lambda dz)dz + (\lambda dz)(z+dz)}{2z\,dz+(dz)^2}\approx\frac{1+\lambda}{2}.$$
Finally, for $u \gt z+dz$, the brown area in the right panel has grown to equal the gray area, whence their ratio is $1$.
These results show that the conditional probability grows linearly from $0$ to $z/(2z)=1/2$ as $u$ grows from $0$ to $z$, then shoots up linearly from $1/2$ to $1$ in the infinitesimal interval between $z$ and $z+dz$, then stays at $1$ for all larger $u$. Here's a graph:
Because $dz$ is infinitesimal, it is no longer possible to distinguish $z$ from $z+dz$ visually: the plot jumps from a height of $1/2$ to $1$.
Putting the foregoing together into a single formula to be applied to any $z$ for which $0 \lt z \le 1$, we could write the conditional distribution function as
$$F_{U|Z=z}(u) = \left\{\begin{array}{ll} 0 & u \le 0 \\ \frac{u}{2z} & 0 \lt u\le z \\ 1 & u \gt z. \end{array} \right.$$
This is a complete and rigorous answer. The jump shows that a probability density function will not adequately describe the conditional distribution at the value $U=z$. At all other points, though, there is a density $f_{U|Z=z}(u)$. It is equal to $0$ for $u\le 0$, $1/(2z)$ for $0 \le u \lt z$ (the derivative of $u/(2z)$ with respect to $u$), and $0$ for $u \gt z$. You could use a "generalized function" to write this in a density-like form. Let $\delta_z$ be the "generalized density" giving a jump of magnitude $1$ at $z$: that is, it's the "density" of an atom of unit probability located at $z$. Then the generalized density at $z$ can be written $\frac{1}{2}\delta_z$ to express the fact that a probability of $1/2$ is concentrated at $z$. In full, we could write
$$f_{U|Z=z}(u) = \left\{\begin{array}{ll} 0 & u \le 0 \\ \frac{1}{2z} & 0 \lt u\lt z \\ \frac{1}{2}\delta_z(u) & u=z \\ 0 & u \gt z. \end{array} \right.$$