What does the sigma notation with i=1 underneath and n on top actually tell me to add?

A sigma with $i=1$ underneath and $n$ on top tells you to let the counter $i$ take each whole number from $1$ to $n$, including $n$, and to add the expression to the right once for each value. The expression is one term written once with a placeholder in it, not the whole sum. It is an abbreviation for an addition, and the only real question is what the counter is doing. Reading the three decorations before reading the term is the whole habit: bottom gives the start, top gives the stop, and the term tells you what is being added.

The order is mechanical. Put $i=1$ into the expression and write the first term; put $i=2$ in and write the second; continue until $i=n$; put plus signs between the copies. For $n=4$ and the term $i^2$ the copies are $1^2$, $2^2$, $3^2$ and $4^2$, and adding gives $1+4+9+16=30$. The top limit is included, so the last term is the one with $i=4$, not the one before it. Nothing is being multiplied by the sigma, and no term has been skipped: four values of $i$ produce four terms. In general the same instruction gives a sum of $n$ terms, with the counter substituted into the term each time.

The counting is where most slips happen. From $1$ to $n$ inclusive is $n$ terms, because the list $1,2,\ldots,n$ has $n$ entries. From $0$ to $n$ inclusive is $n+1$ terms, because nought is a value as well; from $0$ to $4$ there are five terms, not four. From $1$ to $n-1$ is $n-1$ terms. The counter is local: the $i$ under the sigma has no meaning outside that sum, so the same letter can be reused elsewhere without collision. A term that does not mention $i$ is still repeated once for every value; $\sum_{i=1}^{4}1$ is four ones added, not one one.

A term with no counter still gets copied. If the term is $1$ and $n=4$, the sum is $1+1+1+1=4$, not $1$. That is how a sum of ones counts the number of values. The trace also shows compression: $3+6+9+12$ has four terms and each is three times a counter value, so it is the sum of $3i$ from $i=1$ to $4$. Finding the pattern is the work; writing the sigma is only abbreviation.

The natural guess is that the top $n$ means stop before $n$, because many programming loops stop before their upper bound, or that the expression after the sign already describes every term. Neither is right. The sum includes $i=n$, so from $1$ to $n$ there are $n$ terms. With $n=4$ the values are $1,2,3,4$, not $1,2,3$. If the term is $1$, those four values give $1+1+1+1=4$, which is $n$, not $1$. The top number is a stop value, not a count, and the expression is a template, not a total.

A factor's position matters for what is being added. The expression immediately to the right of the sigma is inside the sum. A factor written to the left is outside and applies once, after all the adding. With $x_1=2$, $x_2=4$ and $x_3=6$, the left expression $(1/3)\sum x_i$ gives $(2+4+6)/3=12/3=4$, while the right expression $\sum x_i/i$ gives $2/1+4/2+6/3=2+2+2=6$. The second is not an average, because each term is divided by a different number. Division by $i$ is inside because it mentions the counter; division by $n$ is outside because $n$ does not change as $i$ runs. A factor may be moved across the sign only when it does not mention the counter: $\sum c x_i = c\sum x_i$, but $\sum i x_i$ cannot have $i$ pulled out.

The limits need not be numbers. A sum may run over the members of a set, with the set named underneath and no top limit, and then there is one term per member and no meaningful third term. A condition can replace the bounds. When the top is below the bottom, there are no terms, and the agreed value is nought; a sum from $i=5$ to $4$ is empty and equals $0$. That convention is not an error, it is what lets a recursive formula start at its base case without a special case.