I am fairly new to more abstract mathematical structures and frankly I'm quite new to detailed proofs in general. For context, this is not a homework problem, I am just trying to better understand how to convert conceptual understandings of things into parts of proofs.
I would like to show that the function $$f: \Bbb R[x]/(x^2 - 1) \rightarrow \Bbb R \times \Bbb R $$ $$f(a + bx + (x^2 - 1)) := (a + b, a - b) $$
is surjective. I understand that this requires showing that for every $q \in \Bbb R \times \Bbb R$ there must exist some $p\in \Bbb R[x]/(x^2 - 1) $ such that $f(p)=q$. It seems clear to me that this is the case, because two elements of $\Bbb R$ are being mapped to two elements of $\Bbb R$, but for one thing I don't know if this is always the case, and more importantly that's not even a remotely rigorous explanation.
Assuming I could show that $f$ is a ring homomorphism, how exactly could I show surjectivity? And is there some direction to take for showing this in general or is the route to showing surjectivity dependent on the problem? For instance, I find showing injectivity very straightforward because it only requires one to consider what would cause $a=b$ in the case that $f(a) = f(b)$, which is usually even more straightforward if $f$ is a homomorphism.
First, your statement of what you want to prove is wrong. You want to prove that for each $q\in\mathbb{R}\times\mathbb{R}$, there exists $p\in \Bbb R[x]/(x^2 - 1)$ such that $f(p)=q$. The order of the "for each..." and "there exists..." matters, since the way you wrote it, you pick $p$ before $q$, so you would have the same $p$ for every $q$ which is obviously impossible.
Now, as for how to prove it, just do it! Take an element $q\in\mathbb{R}\times\mathbb{R}$; that is, an ordered pair $q=(c,d)$ where $c,d\in\mathbb{R}$. Now your goal is to find some $p\in \Bbb R[x]/(x^2 - 1)$ such that $f(p)=q$. Looking at the definition of $f$ as $$f(a + bx + (x^2 - 1)) = (a + b, a - b),$$ can you see how you could find $p$ that works?
More details are hidden below.
In general, this is how you prove a map $f:X\to Y$ between two sets is surjective: you take an arbitrary $y\in Y$, and explain how to find an element $x\in X$ such that $f(x)=y$. Really, much more generally, this is how you prove any statement of the form "for all... there exists...": you take an arbitrary case of the "for all...", and explain how to find the thing that needs to exist.
Sometimes, knowing that a map is a homomorphism does make it easier to prove it is surjective. For instance, if $X$ and $Y$ are rings and $f:X\to Y$ is a ring-homomorphism, then the image of $f$ is always a subring of $Y$, and so to show $f$ is surjective, it suffices to show that the image of $f$ contains a set which generates $Y$ as a ring. In other words, you only need to solve $f(x)=y$ for $x$ for a few special values of $y$ which you know are enough to generate all of $Y$. For this particular problem, though, that doesn't really save any substantial amount of work.