I understand basically what a splitting field is and how to obtain it. However, I am unable to fill in the details in the construction. To be concrete,
Let $F$ be a field and $f(x) \in F[x]$ be monic. Write $f(x)=\prod_{i=1}^k f_i(x)$ where each $f_i(x)$ is irreducible. If all the $f_i(x)$ have degree $1$ there's nothing to prove. So suppose $\deg f_1(x) > 1$. We have that $E:=F[x]/(f_1(x))$ is a field, and $r = x+(f_1(x))$ is a root of $f_1(x)$ in $E$. Also $E$ is an extension of the field $F$, so $f(x) \in E[x]$. Writing $f(x)$ as a product of irreducibles in $E[x]$, we see that $f_1(x)$ must decompose into $(x-r) g(x)$ where $g(x)$ is possibly reducible. Hence the number of irreducible factors of $f(x)$ in $E[x]$ is $>k$. So we can proceed by induction on $n-k$, where $n = \deg f(x)$.
Questions:
In what sense is $r$ a root of $f_1(x)$ in $E$? What $f_1(x)$ are we talking about? How to view the elements of $F[x]$ as elements of $E[x]$? i.e., rigorously, given $f(x) \in F[x]$, what kind of $f(x)$ are we talking about when we view $f(x)$ as an element of $E[x]$?
Regarding your first question: You see that $E$ ist a field and so we can form the ring of polynomials over this field. Lets call it $E[T]$, just to (hopefully) minimise confusion. Now, $E$ is a extension of $F$, so we can embedd $F[x]$ into $E[T]$ by just mapping $f(x)$ to $f(T)$. Since the coefficients are in $F$ they are also in $E$, so this is well defined. If we now plug in $x \in E=F[x]/(f(x))$ into the polynomial $f(T) \in E[T]$ we get $f(x)$ back. But $f(x)$ is zero in $E$ because we modded it out. Therefore $x$ is a root in $E$ of $f$.