Proving identities like $\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}$ combinatorially

It seems easier to look at the left-hand side :

$\sum \binom n k ^2 = \binom {2n} n$ is the number of ways to choose $n$ elements out of a set $X$ of $2n$ elements, where we are given a partition $X = X_1 \cup X_2$ with $\# X_i = n$.
Next, $\sum k \binom n k ^2$ is the number of ways to do this and then choose a special element among the ones chosen in $X_1$. So, reversing the order of the choices (choose the special element first), then forgetting as much as possible about the partition, you get that is the number of ways to pick one special element in $X_1$ and then pick $n-1$ other elements in $X$.

Then it shouldn't be too hard to explain why this is equal to the right-hand side.

Since $\binom{n}{k} = \dfrac{n}{k}\binom{n-1}{k-1}$ the identity can be written as $$\sum_{k=1}^n\binom{n}{n-k}\binom{n-1}{k-1} = \binom{2n-1}{n-1}.$$

Suppose we have $2n-1$ items in line and we need to choose $n-1$ items from them. The right hand side is the number of ways to do this in the obvious way. Another way to choose our $n-1$ items from this line: Divide the line into two sections; the first section contains the first $n$ items, the second contains the last $n-1$ items. When picking $n-1$ items from the entire line, we could pick all $n-1$ from the first line and none from the second ($k=1$), or $n-2$ from the first line and $1$ from the second ($k=2$) ... ... or none from the first line and $n-1$ from the second ($k=n$).

For $a,b,n \in \mathbb{N}$ we have

$$\binom{a+b}{n} = \sum_{p+q=n} \binom{a}{p} \cdot \binom{b}{q}$$

since both sides count the subsets of $\{1,\dotsc,a\} \sqcup \{1,\dotsc,b\}$ with $n$ elements. This could also be derived from the isomorphism $\Lambda^n(V \oplus W) = \bigoplus_{p+q=n} \Lambda^p(V) \otimes \Lambda^q(W)$ of exterior powers, where $V,W$ are free modules of rank $a$ resp. $b$ (decategorification).

In particular,

$$\sum_{k=0}^{n} k \binom{n}{k}^2 = n \sum_{k=0}^{n} \binom{n}{n-k} \binom{n-1}{k-1}=n \binom{2n-1}{n-1}=n \binom{2n-1}{n}.$$

One can also derive a lot of other formulas as a special case, for example

$$\sum_{k=0}^{n} \binom{n}{k}^2 = \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k} = \binom{2n}{n}.$$