I am in my first upper-division linear algebra class, and I am confused about the following:
We've proved that $$(-1)a=-a$$ where $a \in \mathbb{F}$. In class, when this came up, professor said "(the additive inverse of 1) $\cdot$ a = (additive inverse of a)". Taking for granted that $-a$ is the additive inverse of $a \in F$, though this gets a bit fuzzy about zero. When subtracting, we also seem to take for granted that $$a + (-b)= a-b$$ I was wondering how we prove this, or if this is simply something we take for experience from $\mathbb{R}$.
Additionally, the statement that in $\mathbb{F}_3=\{0, 1, 2\}$ that $-1 = 2$ is a bit confusing to me. Considering that in $\mathbb{F}_3$ we define addition $a+_{\mathbb{F}_3} b = (a+_{\mathbb{F}}b)\bmod{3}$
$\newcommand{\F}{\mathbb{F}}$ First in $\F_3$ you're working modulo $3$, so if you're sticking exactly to the fact that $-a$ is the additive inverse of $a$, then $-1 \equiv 2$ is for that exact reason. The additive inverse of $1$ in $\F_3$ is two because $$1+2 = 3 \equiv 0 \pmod 3.$$
For the fact that $a+(-b) = a-b$, it can either serve as an axiom or you can think of it in terms of the additive inverses again. I.e $a-b$ is the number such that when $b$ is added to it gives $a$. I.e to show that $a-b = a+(-b)$ it suffices to show that $b+(a+(-b)) = a$.
Edit: You are correct in saying that $-1$ is not an element of $\F_3$ as you understand it. But as Robert Shore mentions, the elements of $\F_3$ are technically equivalence classes to which $-1 \in [2]$. You say you haven't seen these notions, so I stress again the point I left in the comments. As a field, every element has an additive inverse. So when your professor writes $-1 \equiv 2$ he means that the additive inverse of $1$, i.e $-1$, is equal to $2$ in $\F_3$. Compare this with your field axioms, if $a \in F$ it's additive inverse is the element $b$ such that $a+b = b+a =a$. Doesn't $2$ satisfy this property for $1$ in $\F_3$ since $$ 1+2 = 2+1 = 3 \equiv 0 \pmod{3}. $$ This is the same idea as saying $-1\equiv 1$ in $\F_2$. If you're still having trouble I would recommend reading up on modular arithmetic.
More Edits: What you've written as for every $a \in \F$ there exists some $b \in \F$ with $ab = 1_\F$ is not the axiom for an additive inverse, it is the axiom for a multiplicative inverse.
To add another perspective on why we denote the inverses the way we do, one can show that the additive identity of any element of a field is unique. I.e for any $a \in F$ there is only one element $b$ for which $a+b = 0$. Therefore, we are justified in our choice of denoting this unique element by $-a$ because there is only one such inverse. If you've been given the axioms as for every $a \in F$ there exists some $b \in F$ for which $$ a+b = b+a = 0 $$ this is perfectly correct, and you could keep using this notation and mentioning additive inverses all you like, but because these inverses are unique it's certainly more convenient to simply denote it $-a$ and appeal to intuition.