Extensions can be constructed 2 ways to get an extension with roots of a polynomial
Adjoining an element to a field - i.e. $F(\sqrt 2)$ is an extension of $F$. You can also build a tower of extensions by adjoining another element to the first extension.
Extensions generated by a Principal ideal. For e.g. if $p(x)$ is a principal Ideal of $F(x)$, then you can construct an extension $F(x)/\langle p(x) \rangle$. The elements of this extension are cosets.
Though I am able to understand the 2 kinds of constructions, I am unable to understand which constructions is used when/where & if there is any relation between the extensions obtained by the 2 different kinds of constructions?
Let us first try to deal with extensions via an algebraic element.
Let $F$ be a field and $p(x) \in F[x] $ be irreducible and let us ponder about a root of $p(x) $. A very simple approach is to assume that we can apply field operations ($+, -, \times, /$) on the root $x$ and combine it with elements of $F$. However in so doing we always keep track of the fact that it is a root of $$p(x) =a_0x^n+a_1x^{n-1}+\dots+a_n$$ and thus any powers of $x$ greater than or equal to $x^n$ can be replaced with an expression containing powers of $x$ upto $x^{n-1}$ via the equation $$x^n=-\frac{a_1x^{n-1}+\dots+a_n}{a_0}$$ Thus when we try to combine $x$ with elements of $F$ Using $+, -, \times$ we get polynomials in $x$ with coefficients in $F$ and the highest power of $x$ in such polynomials is $x^{n-1}$. Using the fact that $p(x) $ is irreducible one can show that ratio of such polynomials is also a similar expression (without any courses in abstract algebra the proof of this quite non-obvious).
Thus when we try to combine elements of $F$ with the root of $p(x) $ we are naturally led to polynomials in root $x$ with powers upto $x^{n-1}$ and these polynomials behave as elements of a field. Note further that all of this was achieved by replacing $x^n$ and higher powers by using the fact that $x$ is a root of $p(x) $. More abstractly the process can be understood in terms of modulo arithmetic. Let $a(x), b(x) $ be polynomials in $F[x] $ and we write $a(x) \equiv b(x) \pmod {p(x)} $ if $a(x) - b(x) $ is divisible by $p(x) $. This leads us to an equivalence relation on $F[x]$ and a set of equivalence classes say $K$.
Now given a polynomial $a(x) \in F[x] $ let $r(x) $ be the remainder when $a(x) $ is divided by $p(x) $. Then we have $a(x) \equiv r(x) \pmod {p(x)} $ and degree of $r(x) $ is less than $n$ (the degree of $p(x) $). Thus the set $K$ of equivalence classes can be written as $$K=\{[r(x)] \mid\text{degree of }r(x) <n, r(x) \in F[x] \} $$ This is what we denote as $F[x] /\langle p(x)\rangle $ and this set is a field if $p(x) $ is irreducible.
Another point to note here is that there is a subset of $K$ namely $$\{[a] \mid a\in F\} $$ which behaves exactly in the same manner as $F$ with respect to field operations and in this sense we say that $F$ is a subfield of $K$ or that $K$ is an extension of $F$. This is the procedure where one can extend a field by using an irreducible polynomial.
Another approach of "adjoining" assumes there is an extension $K/F$ of fields (obtained using any means, including the one involving an irreducible polynomial described earlier). Let $a\in K$ and the set $F(a) $ is defined to be the intersection of all subfields of $K$ which contain $F$ as well as $a$. Thus it is the smallest subfield containing $F$ and $a$. And it can be proved that it is the set of all expressions of the form $f(a) /g(a) $ where $f(x), g(x) \in F[x] $ and $g(a) \neq 0$.
If an element $a\in K$ is algebraic over $F$ it can be proved that $F(a) $ is isomorphic to $F[x] /\langle p(x)\rangle $ where $p(x) $ is the minimal polynomial for $a$ over $F$. And thus both approaches are same in case the field extension is done via an algebraic element.
Note further that there are other ways to extend a field and most notable among them is the field of fractions of $F[x] $ which leads to $F(x) $, the field of rational functions in an indeterminate $x$. And we have the non-algebraic mechanism of extending from $\mathbb {Q} $ to $\mathbb{R} $ which involves order relations in a fundamental way and does not apply to general fields.