I am currently learning field theory by myself (following Advanced Modern Algebra by Joseph Rotman at Chapter 3).
We suppose $\mathbb{F}$ is some arbitrary field and $P\in\mathbb{F}[x]$, the polynomial ring corresponding to $\mathbb{F}$. Suppose $P$ is irreducible. The Kronecker Theorem for field extensions claims that we can always construct a field $E$ where $P$ splits over $E$. We start by adjoining a root $p_1$ of $P$ to $\mathbb{F}$ to form some field extension $\mathbb{F}_{p_1}$ and I have no problems understanding why $\mathbb{F}[x]/I_{P}$ and $\mathbb{F}_{p_1}$ are isomorphic, where $I_P$ is the ideal generated by $P$ forming a quotient ring for $\mathbb{F}[x]$. This process makes sense if we have a polynomial in $\mathbb{Q}[x]$ since we can adjoin some root in $\mathbb{C}$ to $\mathbb{F}$ since the Fundamental Theorem of Algebra asserts such a root exists. We can pick some roots for quintics using iterative methods and adjoin some particular root to the field. However for some arbitrary collection of objects, together with operations which satisfy the field axioms, how are we defining such a root $p_1$? We know $x+I_p$ is a root of $P$ if we let $P$ be a polynomial with coefficients in $\mathbb{F}[x]/I_P$. But does that guarantee that $p_1$ is well defined, or is $p_1$ merely defined as some arbitrary element not in $\mathbb{F}$ satisfying $P$? I apologise in advance if this question seems elementary or silly or I am thinking in a too abstract or complicated way as I have not started my university education yet (but soon). Please kindly also correct any factual or conceptual errors in any assertions here. I would like to resolve this issue to move on to Galois Theory so I appreciate any help. Thank you.
You have a good understanding (especially since you haven't started university) but you seem a bit muddled, which is understandable when you first deal with roots in more abstract fields.
Proposition
For any irreducible polynomial $f \in \mathbb{K}[x]$, the field $\mathbb{F}=\mathbb{K}[x]/\mathbb{K}[x]f$ is given by $$\mathbb{F} = \mathbb{K}1_{\mathbb{F}} \oplus \mathbb{K}t \oplus \cdots \oplus \mathbb{K}t^{n-1},$$ where $1_{\mathbb{F}} = 1+\mathbb{K}[x]f$ and $t=x+\mathbb{K}[x]f$, and $f(t)=0$.
Definition
Given field extension $\mathbb{K} \subseteq \mathbb{F}$ and $a_1,\dots,a_n$, define $$\mathbb{K}(a_1,\dots,a_n) = \{ \frac{c}{d} \ | \ c,d \in \mathbb{K}[a_1,\dots,a_n], d(a_1,\dots,a_n) \}$$ which is the smallest subfield of $\mathbb{F}$ containing $\mathbb{K}$ and $a_1,\dots,a_n$.
Remark
Now, suppose $f \in \mathbb{K}[x]$ with $\mathrm{deg}f \geq 1$. We can factorise $f$ into irreducibles over $\mathbb{K}[x]$ $f = p_1 \cdots p_r$. Since $p_1 \in \mathbb{K}[x]$ is irreducible, $p_1$ has a root $t_1$ in a larger field $\mathbb{K}_1$. We can factorise $f$ into irreducibles over $\mathbb{K}_{1}[x]$ $f = (x - t_1) q_2 \cdots q_s$. Since $q_2 \in \mathbb{K}_{1}[x]$ is irreducible, $q_2$ has a root $t_2$ in a larger field $\mathbb{K}_2$. We can factorise $f$ into irreducibles over $\mathbb{K}_{2}[x]$ $f = (x -t_1)(x-t_2) h_2 \cdots h_l$.
Continuing like this, we get a larger field $\mathbb{F}$ which splits $f$ into linear factors $f = c (x-t_1) \cdots (x-t_n)$ with $c \in \mathbb{K}$ and $t_1,\dots,t_n \in \mathbb{F}$. That fact that this is possible leads us to the following definition.
Definition
Let $\mathbb{K}$ be a field and let $f \in \mathbb{K}[x] \setminus \{0\}$.
We call a field $\mathbb{F}$, containing $\mathbb{K}$, a splitting field of $f$ over $\mathbb{K}$ if $\exists c \in \mathbb{K}$ $\exists t_1,\dots,t_n \in \mathbb{F}$ such that $f = c (x-t_1) \dots (x-t_n)$ and $\mathbb{F} = \mathbb{K}(t_1,\dots,t_n)$.
Remark
Notice that whilst, given $t_1,\dots,t_n$, $\mathbb{F} = \mathbb{K}(t_1,\dots,t_n)$ is completely and uniquely defined, the $t_1,\dots,t_n$ are definitely not uniquely defined by $f \in \mathbb{K}[x]$.
For example, take $x^2 + 1 \in \mathbb{Q}[x]$. We can say that its roots are $(0,1)$, $(0,-1)$ in $\mathbb{C}$ (as $\mathbb{R}^2$) or we could say that they are $\bar{t}$, $-\bar{t}$ in $\mathbb{R}[t]/\mathbb{R}[t](t^2+1)$.
What we do know is that the roots are unique up to renaming and that therefore $f \in \mathbb{K}[x]$ has only one splitting field up to field isomorphism (i.e. renaming).