To some extent it’s an art that one acquires with practice. What I can do is talk you through my own thoughts on encountering these two problems.
The first one looks as if we’re choosing something out of $[n]$. Specifically, the general term looks as if we’re choosing $k$ things and then designating two of them (because the $k$ is squared) as special. For instance, we might have a pool of $n$ candidates from which we want to choose a committee of $k$, make one member of the committee the chairperson, and make one member, possibly the same one, the secretary. We’re summing over all possible $k$, so we should get the number of ways to pick a committee of arbitrary size from our pool of $n$ candidates, choose a chairperson, and choose a secretary (who might be the same as the chairperson). Can we interpret the righthand side as counting the same thing more directly?
The first term is clearly the number of ways to pick the chairperson, pick a different person as secretary, and then fill out the committee with any subset of the remaining $n-2$ people: $n(n-1)2^{n-2}$ is the number of committees whose chairperson and secretary are different. The second term is the number of ways to pick one of the $n$ people to be both chairperson and secretary and then to pad out the committee with any subset of the remaining $n-1$ people: there are $n2^{n-1}$ possible committees on which the chairperson and the secretary are the same person. These two subsets of the committes are exhaustive and don’t overlap, so their sum, $n(n-1)2^{n-2}+n2^{n-1}$ is indeed the total number of possible committees, and the identity is established.
Let’s see if a similar approach works on the second identity. If it were $2n\binom{2n-1}{n-1}$, we could interpret it as the number of ways of picking a captain from a pool of $2n$ players and then filling out a team of $n$ by picking $n-1$ more players from the remaining $2n-1$. We have only half of this, so perhaps we’re picking the captain from a sub-pool of $n$ of the players. Let’s say that we have $n$ women and $n$ men, and we want to know in how many ways we can pick a team of $n$ that has a woman as captain. There are $n$ ways to choose the captain from the pool of $n$ women, and there are then $\binom{2n-1}{n-1}$ ways to choose the rest of the team, so $n\binom{2n-1}{n-1}$ does indeed give the correct count. Now the challenge is to interpret the lefthand side as counting the same thing.
For each $k$ we seem to be choosing $k$ members of one set of $n$ (one factor of $\binom{n}k$), $k$ members of another set of $n$ (another factor of $\binom{n}k$), and then singling one of (say) the first $k$ members (the factor of $k$). For instance, we might be choosing $k$ of the $n$ women and $k$ of the $n$ men and then picking one of the $k$ chosen women to be captain. The problem is that this gives us a team of $2k$ people, not a team of $n$. But we do get a team of $n$ if we interpret $\binom{n}k^2$ as the number of ways to choose $k$ women to be on the team and $k$ men who will not be on the team: the remaining $n-k$ men will be on the team, and we’ll have a team of $n$. (In more algebraic terms we’re just noticing that $\binom{n}k^2=\binom{n}k\binom{n}{n-k}$.) Thus, the $k$ term on the lefthand side is the number of ways to choose a team of $n$ that has $k$ women, one of whom is the captain. Summing over $k$ clearly gives the desired result.
