Let's consider the polynomial ring $R=F[x]$ over a field $F$. Then by taking the quotient by a principal ideal $I=(f(x))$ generated by an irreducible polynomial $f(x)$, we obtain a field $R'=F[x]/(f(x))$.
It's easy to see that $R'$ is indeed a field. Since the ideals of $R$ which contain $I$ are in bijective correspondence with the ideals of $R'$, we can conclude that $R'$ has only two ideals and is therefore a field (as $I$ is maximal in $R$ since $f(x)$ is irreducible).
I wanted to ask, is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field? I would ideally like some way of demonstrating that the existence of a nonzero polynomial equivalent to zero in $R'$ somehow allows us to describe an algorithm to calculate multiplicative inverses...
This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?
The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :
This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x \notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) \in R \times I$ such that $ax+j=1$ or $\bar a \bar x = \bar 1$ in $R/J$ proving that $R/J$ is a field.
Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.
And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.
Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.