Proof attempt: Since f is irreducible and has a as root then K(a) is isomorphic to \frac{K[x]}{(f

Step 1
For the first question, we just need to check the last condition ${\displaystyle w\left(a\right)=b}$. But just looking at the definition of the two isomorphisms in the composite
${\displaystyle w:K\left(a\right)\to \frac{K\left[x\right]}{\left(f\right)}\to K\left(b\right)}$
the first maps a to ${\displaystyle x+\left(f\right)}$, and the second maps ${\displaystyle x+\left(f\right)}$ to b. So ${\displaystyle w\left(a\right)=b}$, as we desire.
For the second question, we start by considering the evaluation K-algebra homomorphism
${\displaystyle \varphi :K\left[x\right]\to K\left(a\right)}$
which maps x to a.
Then since ${\displaystyle f\left(a\right)=0}$, every ${\displaystyle g\in \left(f\right)}$ is mapped to zero by ${\displaystyle \varphi }$. (Just note that ${\displaystyle \varphi \left(g\right)=g\left(a\right)}$.) Thus ${\displaystyle \varphi }$ descends to a K-algebra homomorphism
$\tilde{\varphi }:\frac{K[x]}{(f)}\to K(a)$.
To show that
${\displaystyle K\left(a\right)\stackrel{\sim }{=}\frac{K\left[x\right]}{\left(f\right)}}$,
it just remains to show that
$\tilde{\varphi }:\frac{K[x]}{(f)}\to K(a)$
is an isomorphism. Indeed, being a map from a field, $\tilde{\varphi }$ is automatically injective.
Step 2
Finally, to check that $\tilde{\varphi }$ is surjective, we just need a characterization of an arbitrary element of ${\displaystyle K\left(a\right)}$. We'll use that
${\displaystyle K\left(a\right)=K\left[a\right]}$,
which is proved in many places (including I'm sure on this site). Given this there is essentially nothing to do: any element v of ${\displaystyle K\left[a\right]}$ is just a polynomial in a with cofficients in K, say
${\displaystyle v=3a^2+42a^3+6}$,
and replacing each occurrence of a with x gives a polynomial (e.g. ${\displaystyle 3x^2+42x^3+6}$) which is mapped to v by ${\displaystyle widetilde\left\{\varphi \right\}}$. This completes the proof.
Step 4
You might ask: ''Hang on, this seems to make sense but how did we use anywhere that f was a minimal polynomial?''. Indeed, so long as f was any polynomial over K with the root a we could still construct the map
$\tilde{\varphi }:\frac{K[x]}{(f)}\to K(a)$
and see that this map is surjective. The problem is that we lose injectivity: f will in general of course not be irreducible (equivalently, up to a scalar the minimal polynomial of a), so ${\displaystyle \frac{K\left[x\right]}{\left(f\right)}}$ won't be a field.

Step 1
To expand on lhf's comments: the polynomial ring ${\displaystyle K\left[X\right]}$ has the property that for any commutative ring extension of the form ${\displaystyle K\left[\alpha \right]}$, there is a unique homomorphism
${\displaystyle K\left[X\right]\to K\left[\alpha \right]}$
fixing K and sending ${\displaystyle X\mapsto \alpha }$. This is just a general property of polynomial rings. Now, if ${\displaystyle \alpha }$ is algebraic over K, then the kernel of this homomorphism is the ideal ${\displaystyle \left(f\alpha \right)}$ generated by its minimal polynomial, and by the fundamental theorem on homomorphisms we then have
${\displaystyle K\left[\alpha \right]\stackrel{\sim }{=}\frac{K\left[X\right]}{\left(f\alpha \right)}}$,
and the isomorphism fixes K and sends ${\displaystyle \alpha \mapsto \bar{X}}$
Now if a,b are algebraic with the same minimal polynomial f, the corresponding ring extensions ${\displaystyle K\left[a\right]}$ and ${\displaystyle K\left[b\right]}$ are isomorphic to ${\displaystyle \frac{K\left[X\right]}{\left(f\right)}}$, both via an isomorphism sending ${\displaystyle a\mapsto \bar{X}}$ or ${\displaystyle b\mapsto \bar{X}}$ and fixing K. Now just compose one of the isomorphisms with the inverse of the other one to get an isomorphism between ${\displaystyle K\left[a\right]}$ and ${\displaystyle K\left[b\right]}$ which fixes K and maps ${\displaystyle a\mapsto b}$
Now finally note that ${\displaystyle K\left[\alpha \right]=K\left(\alpha \right)}$ for algebraic ${\displaystyle \alpha }$, so the isomorphism also applies to the field extensions in question.

Step 1 Consider the homomorphism $K[X]\to L$ $P(X)\mapsto P(a)$ Its image is $K(a)$ and its kernel is the ideal (f), whence, by the First Isomorphism theorem, $\frac{K[X]}{(f)}\cong K(a)$ and similarly for $K(b)$.