We roll a fair die until we get $6$ for the second time. Let $X_1$ be the number of rolls until we get $6$ for the first time, and let $X_2$ be the number of rolls until we get $6$ for the second time. Find the conditional expectation of $X_1$ given that $X_2 = 5$.
Solution: Here the probabilities for $X_1$ to take any of the $4$ values are the same. For example, if $X_1 = k$, this means that at the step with number $k$ $6$ appeared, and at the remaining steps from $1$ to $4$, not $6$ appeared. The probability is $\frac{1}{6}\frac{5}{6}^3$ regardless $k$. Therefore, the average value of $X_1$ is $\frac{1 + 2 + 3 + 4}{4} = \frac{5}{2}$.
But I am not sure about this solution, mostly because there is "conditional expectation" in the question, which must be found using standard approach conditional expectation.
Under condition $X_2=5$ random variable $X_1$ only takes values in $\{1,2,3,4\}$ and it is perfectly legal to deduce $\mathbb E[X_1\mid X_2=5]$ by means of:$$\mathbb E[X_1\mid X_2=5]=\sum_{k=1}^4kP(X_1=k\mid X_2=5)\tag1$$
As you observed $P(X_1=k\mid X_2=5)$ does not depend on $k$ so that it must take value $\frac14$ for every $k\in\{1,2,3,4\}$ and consequently: $$\mathbb E[X_1\mid X_2=5]=\frac14\sum_{k=1}^4k=\frac52$$
In the link that you added you can find a classical definition:$$\mathbb E[X\mid H]=\int xdP(x\mid H)\tag2$$where $X$ denotes a random variable and $H$ denotes an event.
That is the situation that we are in here with $X_1$ as random variable and $\{X_2=5\}$ as event.
Actually $(1)$ is just a work out of $(2)$.