I guess it is quite easy to see that $\frac{\mathbb{F}[x]}{(f(x))} = \{c_1 + c_2x + c_3x^2 + \ldots + c_nx^{n-1} \mid c_i \in \mathbb{F}\}$ and thus you have $n$ coefficients which you can all map to one of the $n$ unity vectors in $\mathbb{F}^n$. The only condition for this is that you have $\deg(f(x)) = n$.
If you take a look here, you might find what you need under "Statement for general rings", but I tried proving it without trusting wikipedia this way (as an exercise).
The Chinese remainder theorem for elements of $\mathbb{F}[x]$:
If $\forall i,j : \gcd(x-a_i, x-a_j) = 1$, we have $\forall b_1(x), \ldots, b_n(x) \in \mathbb{F}[x]: \exists c(x) \in \mathbb{F}[x]$ so that
$$\left\{\begin{align} c(x) & \equiv b_1(x) \mod{x - a_1} \\ c(x) & \equiv b_2(x) \mod{x - a_2} \\ & \vdots \\ c(x) & \equiv b_n(x) \mod{x - a_n} \end{align}\right.$$
and this solution is unique mod $f(x)$.
If you now take $b_i(x) \in \frac{\mathbb{F}}{(x-a_i)}$, then you can map $c(x)$ on $(b_1(x), b_2(x), \ldots, b_n(x))$, which is a mapping from $\frac{\mathbb{F}[x]}{(f(x))}$ to $\frac{\mathbb{F}[x]}{(x-a_1)} \times \frac{\mathbb{F}[x]}{(x-a_2)} \times \ldots \times \frac{\mathbb{F}[x]}{(x-a_n)}$
More formally you could define $\left(\text{with }[g(x)]_{k(x)} = \{g(x) + q(x)k(x) \mid q(x) \in \mathbb{F}[x]\}\right)$:
$$\begin{align} f:\frac{\mathbb{F}[x]}{(f(x))} & \to \frac{\mathbb{F}[x]}{(x-a_1)} \times \frac{\mathbb{F}[x]}{(x-a_2)} \times \ldots \times \frac{\mathbb{F}[x]}{(x-a_n)} : \\
[g(x)]_{f(x)} & \mapsto f([g(x)]_{f(x)}) = ([g(x)]_{x-a_1}, [g(x)]_{x-a_2}, \ldots, [g(x)]_{x-a_n}) \end{align}$$
This is a homomorphism of rings because $f([g(x) + h(x)]_{f(x)}) = f([g(x)]_{f(x)}) + f([h(x)]_{f(x)})$, $f([g(x)h(x)]_{f(x)}) = f([g(x)]_{f(x)}) * f([h(x)]_{f(x)})$ (with * the pairwise multiplication) and $f([1]_{f(x)}) = (1,1,\ldots,1)$ (NOTE: check this yourself). Because $f$ is a bijection (this can easily be shown when you calculate the function value for $g_i(x) = \frac{f(x)}{x-a_i}$), $f$ is an isomorphism and we have that $\frac{\mathbb{F}[x]}{(f(x))} \cong \frac{\mathbb{F}[x]}{(x-a_1)} \times \frac{\mathbb{F}[x]}{(x-a_2)} \times \ldots \times \frac{\mathbb{F}[x]}{(x-a_n)}$. If you take the argument from the beginning from the answer to prove that $\frac{\mathbb{F}[x]}{(x-a_i)} \cong \mathbb{F}$, you can state that $\frac{\mathbb{F}[x]}{(f(x))} \cong \mathbb{F}^n$.
Correct me if I made a mistake somewhere or if something isn't clear, please.
Thanks for your attention.
