I would like to prove $\sum_{k=0}^{n}{k {n \choose k}}=n2^{n-1}$ with a combinatorial proof, once I already know to prove it algebrically. I thought about it in the following way:
Consider a set $S=\{x_1,...,x_n\}$ with $n$ elements. How many ways can I choose a subset of S with a leader element?
I) Consider a subset of $S$ with $k+1$ elements. I can start by choosing the $k$ non-leader elements ,${n \choose k}$, and then from the other $n-k$ elements I can choose the leader, ${n-k \choose 1}$. Therefore there exist ${n-k \choose 1}{n \choose k}=(n-k){n \choose k}=(n-k){n \choose n-k}$ ways of choosing a subset of $S$ with $k+1$ elements. Therefore there are $\sum_{k=0}^{n-1}{(n-k) {n \choose n-k}}$.
II)Every element can be leader. Therefore for each leader I have to join a subset of $S$ that does not include the leader, $n2^{n-1}$.
Is it a combinatorial proof? I would be very happy if anyone could share a different one.
Your reasoning is correct, but what it proves is that
$$\sum_{k=0}^{n-1}(n-k)\binom{n}{n-k}=n2^{n-1}\,;$$
in order to complete the proof, you have to show that
$$\sum_{k=0}^{n-1}(n-k)\binom{n}{n-k}=\sum_{k=0}^nk\binom{n}k\,.$$
This isn’t hard:
$$\sum_{k=0}^{n-1}(n-k)\binom{n}{n-k}=\sum_{k=1}^nk\binom{n}k=\sum_{k=0}^nk\binom{n}k\,.$$
But you can avoid this step by using the story suggested in the comments: $k\binom{n}k$ is the number of ways to choose $k$ elements and then choose one of those $k$ to be the leader.