Let $K$ be an algebraically closed field and suppose that $f \in K[x,y]$ has decomposition in irreducible factors given by $f=f_1^{n_1} \dots f_s^{n_s}$. It is safe to say that every factor $f_i$ has the form $a_ix+b_iy+c_i$ where $a_i,b_i,c_i \in K$ ?
I understand that in one variable, it is possible to write
$f(x)=c(x-a_1)^{n_1}\dots(x-a_s)^{n_s}.$
My question is: If $f \in K[x,y]$ has decomposition in irreducible factors given by $f=f_1^{n_1} \dots f_s^{n_s}$, it is possible do deduce a similar formula?
No, there exists irreducible polynomials which do not have degree 1. Such as $y-x^2$ mentioned in the comments. To see why this is the case use Hilbert's Nullstellensatz (here we need the algebraic closure of $K$) to get a correspondence
$$ f \text{ is irreducible } \iff (f) \text{ is a prime ideal } $$
and note that $$ K[x,y]/(y-x^2) \cong K[x,x^2] \cong K[x]$$ which is an integral domain$ \implies (y-x^2)$ is prime.
As for a general form comparable to $f(x) = c(x-a_1)^{n_1}...(x-a_s)^{n_s}$, I'm not sure the following is what you are looking for but I think it is at least worth mentioning.
Recall that Hilbert Nullstellensatz states that the maximal ideals of the ring $K[x,y]$ are of the form $(x-a_1,y-a_2)$. In particular for any polynomials $f$ there exists a root $(a_1,a_2) \in K^2$ of $f$ such that. $$ f \in (f) \subset (x-a_1,y-a_2).$$ That is $$f = h_1(x-a_1) + h_2(y-a_2)$$ for some polynomials $h_1,h_2 \in K[x,y]$.
In general when working on an algebraically closed field polynomials are $K[x_1,...,x_n]$-linear combinations of the monomials $(x_i-a_i)$ where $(a_1,...,a_n)$ is a root of $f$. In particular if $f = f_1^{n_1}...f_s^{n_s}$ then find a root of each $f_i$ and write $f_i$ as an $K[x_1,...,x_n]$-linear combination of the monomials corresponding to that root.