"Polynomial Division" when coefficients are formal power series?

Question

I read the following Theorem:

Let $ F(x,y) \in K[x,y] $ with $ F(0,0) = 0 $. Let $ \frac{ \partial F}{\partial y} \neq 0 $ at $ (0,0) $. Then there is a unique $ f(x) = c_1 x + c_2 x^2 + ... $ in $ K[[x]] $ such that $$ F(x, c_1x + c_2 x^2 + ... ) = 0 $$

and the proof that was given was actually very constructive in finding those $ c_i $. However, after the Proof there was a Remark, that:

Since $ y = c_1x + c_2 x^2 + ... $ is as Solution in $ K[[x]] $ of the equation $ F(x,y) = 0 $, it follows from the [above] Theorem, that $$ F(x,y) = [y - (c_1x + c_2x^2 + ...) ] F_1(x,y) $$ with $ F_1(x,y) \in K[[x]][y] $.

What I now don't understand is, how this factorization follows from the 1. Theorem? Is it bad wording or do I miss something important here? From my intuition it looks like that:

We interpret our Polynomial $ F(x,y) \in K[x,y] $ as a polynomial in $ y $ with coefficients in $ K[[x]] $. Having that perspective we can call $ y = c_1x + c_2x^2 + ... $ a root (lying in $K[[x]]$) of our polynomial, which is not a following from a theorem but the construction of our power series ("since $ y = $ ...").

Now having a root of our polynomial we can take $ (y - (c_1x + c_2x^2 + ...)) $ as a linear factor of our polynomial. Then we can calculate our other factor $ F_1(x,y) $ via polynomial division.

Am I missing something here? Is it natural that we can do polynomial division when having $ K[[x]] $ as our coefficients (It's still a ring though)? Do we actually not need the theorem for the factorization (given that $ y = ...$ satisfy $f(x,y) = 0$)? Is there something tricky I have to watch out for?

I'm quite unsure if I missed anything or my thoughts are actually correct.

For completeness: $ K $ is an algebraically closed field.