This problem is in NP, because it is "exactly as hard" as SUBSET-SUM problem. Moreover, it is, in some sense, the SUBSET-SUM problem in disguise! Let's show that.
Denote the number of buckets by $n$, their capacities by $c_1 \leqslant c_2 \leqslant \ldots \leqslant c_n$ and the original quantities of water in them by $a_1, a_2, \ldots, a_n$ (here I assume without loss of generality that the buckets are sorted by their capacities). I claim that the following statements are equivalent:
- There exists a way to get exactly $m \in [0, c_n]$ water in some bucket.
- There exists a subset $S$ of $\{1, 2, \ldots, n\}$, such that $\sum\limits_{i \in S} a_i \equiv m \pmod{\gcd(c_1, c_2, \ldots, c_n)}$.
To prove the $1 \Rightarrow 2$ implication, let's notice that, at every moment of time, the quantity of water in each bucket can be expressed as "sum of $a_i$ over some subset of $\{1, 2, \ldots, n\}$" $+$ "some integer linear combination of $c_i$, with negative coefficients allowed". Indeed, we can check that all three allowed operations preserve this property. The latter part is divisible by $\gcd(c_1, c_2, \ldots, c_n)$, hence $m$ being possible to obtain implies the condition $2$.
The $2 \Rightarrow 1$ implication follows from the same type of the argument as I presented in the comments. I will reproduce it here for completeness. Indeed, suppose that $m \equiv \sum\limits_{i \in S} a_i \pmod{\gcd(c_1, c_2, \ldots, c_n)}$. Because $\gcd(c_1, c_2, \ldots, c_n)$ is representable as a linear combination of $c_i$, we can represent $m$ as an integer linear combination of $c_i$ and $a_i$, with all coefficients before $a_i$ being $0$ or $1$.
Clearly, it is enough to obtain $m \bmod c_1$ water somehow (after doing that, we can just repeatedly add $c_1$ of water). But, as long as we are interested in obtaining something only modulo $c_1$, we can replace all negative coefficients before $c_i$ in this sum with nonnegative ones (by adding $c_1$ appropriate number of times). Hence, we have a representation $m \equiv \sum\limits_{i \in S} a_i + \sum\limits_{i = 2}^n k_i c_i \pmod{c_1}$, where $k_i$ are some nonnegative integers.
Now, suppose that we have $x$ water in the $i$-th bucket and $y$ water in the first bucket. Then, we can get $(x + y) \bmod c_1$ water in the first bucket and make the $i$-th bucket empty. Indeed, try to move all the water from $i$-th bucket to the first. If $x + y < c_1$, then we are done. Otherwise, the first bucket is full now and the $i$-th bucket contains $x + y - c_1$ water. Empty the first bucket and repeatedly drain $c_1$ of water from the $i$-th bucket, until there is only $(x + y) \bmod c_1$ water left. Then move all the water to the first bucket.
By doing the operation from the previous paragraph for all $i$ in $S$, we can get $\left( \sum\limits_{i \in S} a_i \right) \bmod c_1$ water in the first bucket (if $1$ is not in S, we need to empty the first bucket beforehand). Now, for each $i$ from $2$ to $n$, repeat the same operation $k_i$ times, but now fill the $i$-th bucket before the operation. In the end, the first bucket will contain exactly $\left( \sum\limits_{i \in S} a_i + \sum\limits_{i=2}^n k_i c_i \right) \bmod c_1 = m \bmod c_1$ water.
Hence, this problem is equivalent to some kind of "modular subset sum", with modulo being equal to $\gcd(c_1, c_2, \ldots, c_n)$. As for NP witness, any witness for "modular subset sum" problem will do, for example the set $S$.
A more natural witness in the context of the original problem is a "compressed sequence of operations", in the sense that we allow commands of the type "repeat a given sequence of operations a given number of times". The proof above can be translated into such a witness, and a particularly simple one (no nested block commands, we repeat only some fixed short blocks, et cetera). I am not 100% sure whether it is possible to simulate a general compressed sequence of operations in polynomial time, but I am inclined to believe that the answer is "yes, it is possible". However, it is definitely possible to simulate sequences arising from the above proof quickly, though requiring that makes the witness much less natural.
Indeed, the second part of the proof constructs a sequence of operations of polynomial length, but with a catch that the following two "mega-operations" are also allowed:
- Given a bucket with $x$ water, repeatedly drain it by $c_1$, till there is only $x \bmod c_1$ water left. The first (the smallest) bucket should be empty before this operation. This operation is used repeatedly thorough the proof
- If we have $m \mod c_1$ water in the first bucket, empty the largest bucket, move $m \mod c_1$ water into it and repeatedly add $c_1$ water until there is $m$ water in the largest bucket. Used only once in the end.
These "mega-operations" correspond to some simple arithmetical operations on the quantities of waters in the buckets and $c_i$'s, hence they can be carried quickly in the model of computation we are using, despite the fact that they take very large number of elementary operations with buckets.
