Equals. $k[[t]]$ is a unique factorization domain, so $k[[t]][x]$ is a unique factorization domain. Thus we can assume that $g$ and $f$ share no common factors. Let $p$ be a prime dividing $g$ that is not a unit mod $t$. Then since its Newton polygon is flat, and the coefficient of some power of $x$ is nonzero mod $t$, the Newton polygon must be horizontal, so the leading coefficient must be nonzero mod $t$.

We know that $p$ divides $f$ in $k[x][[t]]$, so $pa=f$ for some $a$. We will show that $a\in k[[t]][x]$. Assume not, then the degree of $x$ in $a$ at different powers of $t$ is unbounded. Let $n$ be the first power of $t$ where it gets above the degree of $f$ minus the degree of $p$. Modulo $t^{n+1}$, $p$, $a$, and $f$ are all polynomials in $\left(k[t]/t^{n+1}\right)[x]$, and $pa=f$. The degree of $p$ plus the degree of $a$ is greater than the degree of $f$, so the leading coefficients of $p$ and $a$ are zero divisors. But the leading coefficient of $p$ is a unit in $k[t]/t^{n+1}$, so this is impossible. This is a contradiction, so $a\in k[[t]][x]$, so $p|f$ in that ring, so $g$ and $f$ share a common factor, a contradiction.

Therefore, all primes dividing $g$ are units mod $t$, so $g$ is a unit mod $t$, so it is congruent to an element of $k$ modulo $t$.