Combinatorial proof of $\sum\limits_{k=0}^n {n \choose k}3^k=4^n$

Yes, I agree with your interpretation of the left side, and also lhf's comment can be seen in the same way:

1. $4^n$ the ways of divide $n$ balls in $4$ boxes
2. $(3+1)^n$ the same as above
3. $ \sum_{k=0}^n {n \choose k} 1^{n-k} 3^k$ for every $k$, the ways to choose $k$ balls among the $n$ balls you have, times the ways to put $n-k$ balls in a box, times the ways to put the remaining $k$ balls in the remaining $3$ boxes
4. $\sum_{k=0}^n {n \choose k}3^k$ as above, using $1^{n-k}=1$

The RHS looks like the formula for the number of base-4 strings. So we can also interpret the LHS similarly to the balls counting as counting all possible base-3 strings that can be placed in the $n$ length string and filling the rest with the remaining digit.

For example

$$XXX012201XX021XX$$ where $\binom n k$ chooses where $0,1,2$ go and $3^k$ counts every possible string.

You should be careful when specifying buckets since they are distinct, which is clear from the ordering of a string.