Suppose that $f$ has degree $m$ and $g$ has degree $n$, say $ f = a_0y^m + a_1y^{m-1} + \cdots + a_m$ $g = b_0y^n + b_1y^{n-1}+ \cdots + b_n$ Then the resultant of $f$ and $g$ is the determinant of the $(n + m)\times (n+m)$ Sylvester matrix:
$Res(f,g) = \det \left(\begin{matrix} a_0 & a_1 & \cdots & a_m & 0 &\cdots & 0\\ 0 & a_0 & a_1 & \cdots & a_m &\cdots & 0\\ & & & \ddots\\ 0 & \cdots & 0 & a_0 & a_1 &\cdots & a_m \\ b_0 & b_1 &\cdots & b_n &0 &\cdots & 0\\ 0 &b_0 &b_1&\cdots & b_n &\cdots & 0\\ & & & \ddots\\ 0 & \cdots & 0 & b_0 & b_1 & \cdots &b_n \end{matrix}\right).$ To make notation easy, let $A$ denote this matrix. Note that $A\left(\begin{matrix} y^{n+m-1}\\ y^{n+m-2}\\\vdots\\ y\\ 1\end{matrix}\right) = \left(\begin{matrix}y^{n-1}f\\ y^{n-2} f\\\vdots\\ yf\\f\\y^{m-1}g\\y^{m-2}g\\\vdots\\yg\\ g\end{matrix}\right).$ If $B$ is the adjugate matrix of $A$, that is, $BA = \det(A)\,Id = Res(f,g)\,Id$, then multiplying both sides of this equality on the left by $B$ gives $ Res(f,g)\left(\begin{matrix} y^{n+m-1}\\ y^{n+m-2}\\\vdots\\ y\\ 1\end{matrix}\right) = B \left(\begin{matrix}y^{n-1}f\\ y^{n-2} f\\\vdots\\ yf\\f\\y^{m-1}g\\y^{m-2}g\\\vdots\\yg\\ g\end{matrix}\right).$ If we then look in particular at the bottom row of this equality, you get an expression of the form $Res(f,g) = uf + vg,$ as desired.