$P(X<2Y)$ when both $X$ and $Y$ are uniformly distributed $[0,1]$

Clement's answer is completely correct and the more rigorous way of approaching this problem. However I think it is worth noting that in many cases, it can often be useful to draw out a portion of the problem to get an idea of what is going on.

The graph above is just the highlighted areas of where the statement is true. We know that if $Y\geq.5$ then any value of $X$ will work. This is expressed by the blue box. We then need to start caring what situations where its possible to have $X>2Y$. We realize that this is just the line $y=.5x$ and is expressed by the red line. The highlighted area is now where the condition is met in this situation.

Simply finding the area of these shapes results in $\dfrac{1}{2}$ and $\dfrac{1}{4}$ respectively. If you notice Clement's answer, these are the values of the two integrals they found! This should make sense as the double integral is just finding the area of these shapes!

This is a more visual approach to the problem and not ideal for always solving problems of this type, but certainly a worth while intuition check in certain situations.

The issue with your attempt is that the upper bound for the inner integral has to be $\min(1,2y)$, not $2y$ (as $x\leq 1$ a.s., but one can have $2y>1$).

$$ \int_0^1 \int_0^1 f(x) f(y) \mathbf{1}_{x<2y}dx dy = \int_0^1 \int_0^1 \mathbf{1}_{x<2y}dx dy = \int_0^1 \int_0^{\min(2y,1)} dx dy = \int_0^1 dy \int_0^{\min(2y,1)} dx $$ Now, to compute this, you can split the outer integral as $\int_0^{1/2}+\int_{1/2}^1$.

$$ \int_0^1 dy \int_0^{\min(2y,1)} dx = \int_0^{1/2} dy \int_0^{2y} dx + \int_{1/2}^{1} dy \int_0^{1} dx = \int_0^{1/2} 2y dy + \frac{1}{2}\cdot 1 = \frac{1}{4}+\frac{1}{2} $$ and the answer is $\frac{3}{4}$.