My attempt
$$E[U_1 | U_1 > U_2] = E[U_1 | U_1 - U_2 > 0] = \int_{0}^{1} u_1 f_{U_1|U_1-U_2}(a|b) du_1$$
The conditional density function is,
$$f_{U_1|U_1-U_2}(a|b) = \frac{f_{U_1,U_1-U_2}(a,b)}{f_{U_1-U_2}(b)}$$
So there are two things that I need to figure out. One is the joint distribution of $U_1$ and $U_1-U_2$ and the second is the marginal distribution of $U_1-U_2$. I know the latter is the triangular distribution from the convolution formula.
Where I'm having trouble is the joint distribution in the numerator.
My strategy is to first calculate the joint cdf and then take the derivative.
$$F_{U_1,U_1-U_2}(a,b) = P(U_1 \le a, U_1 - U_2 \le b) = P(U_1 \le a, - U_2 \le b - a) $$
If there was no negative in front of $U_2$ then I feel I could set up the double integral.
Motivation
I actually know the answer is $\frac{2}{3}$ because I know how to calculate $E[|U_1 - U_2|] = \frac{1}{3}$ which is a more straight forward problem to me, and then by using conditioning on $E[|U_1 - U_2|]$ get the answer to $E[U_1|U_1>U_2]$ which you can see here.
However, for this question I'm interested in how to directly calculate $E[U_1|U_1>U_2]$ using the definition of conditional expectation. All the stuff I've learned so far is $E[X|Y=y]$, that is conditional expectation where $Y$ is equal to a specific value, versus something like $X>Y$. Thanks for your help.
You are not conditioning over a random variable (or its sigma algebra), but rather over an event.
By definition for conditioning over a non-zero probability event:$$\begin{align}\mathsf E(U_1\mid U_1 > U_2) = &~\dfrac{\mathsf E(U_1\mathbf 1_{U_1>U_2})}{\mathsf P(U_1>U_2)}\\ =&~ \dfrac{\displaystyle \iint_{u>v} u~f_{U_1,U_2}(u,v)~\mathsf d (u, v)}{\displaystyle \iint_{u>v} f_{U_1,U_2}(u,v)~\mathsf d (u, v)}\\ =&~ \dfrac{\displaystyle \int_0^1\int_v^1 u~\mathsf d u~\mathsf d v}{\displaystyle \int_0^1\int_v^1 1~\mathsf d u~\mathsf d v}\\ \vdots~&\end{align}$$