Upper limit of summation index lower than lower limit?

The sum $$\sum_{k=a}^b f(k)$$ for integers $a$ and $b$ is usually (often, but not exclusively) interpreted as the sum of all values of $f(k)$ for $k$ an integer, $a\leq k\leq b$.

If $a\gt b$, then there are no integers $k$ that satisfy the condition $a\leq k\leq b$, so this is a sum with no summands (or "empty sum").

In order to make sure that associativity laws are respected, empty sums (sums with no summands) are defined to be equal to $0$ (likewise, empty products are defined to be equal to $1$).

Since your sum is empty, it is equal to zero by definition.

See also: this question (discussing the value of a "product with no factors"); the same argument for why the product with no factors "should" be equal to $1$ applies to see why the sum with no summands "should" be equal to $0$, based on associativity.

Mathematica treats a sum $\sum_{k=a}^{b}f(k)$ as the sum $f(a)+f(a+1)+\cdots + f(a+\lfloor{b-a}\rfloor)$ which is one possible interpretation that allows sums with non-integer arguments. You can write Sum[f[k],{k,0,-1,-1}] to get the result of the comment.

Another common interpretation is that this sum runs over all integers between $a$ and $b$.

Another common and useful interpretation is the one that for integers $a,b,c$ respects $$\sum_{k=a}^{b-1}f(k) +\sum_{k=b}^{c-1}f(k) =\sum_{k=a}^{c-1}f(k).$$

This interpretation has the big advantage that many summation formulas hold for positive and negative upper summation index simultaneously.

For your sum, all common interpretations give $0$ even though the last interpretation gives $\sum_{k=0}^{-2}f(k) = -f(-1)$

I think it is a convention that summations start at the lower limit and increment by $1$ until the upper bound is reached. In your example there is no allowable value for $k$ that is no larger than the upper bound. You are adding together an empty collection of elements. Suppose that $A$ is a set of numbers. Consider

$$N = \sum A$$ and

$$ \sum (A \cup \varnothing) = \sum A + \sum \varnothing.$$

Since $A = A \cup \varnothing$ the two sums should be equal. But then $\sum \varnothing = 0$.