Perhaps you will find the following helpful, which expands a bit on Robert Israel's answer: Write $f(x)=f_0x^m+\ldots+f_m=f_0\cdot(x-\alpha_1)\cdots(x-\alpha_m).$ Multiplying out the RHS we get $f(x)=f_0\cdot \sum_{j=0}^m(-1)^j\sigma_j(\alpha_1,\ldots,\alpha_m)\cdot x^{m-j}$ where $\sigma_j=\sigma_j(\alpha_1,\ldots,\alpha_m)$ is the $j^{th}$ elementary symmetric polynomial in $\alpha_1,\ldots,\alpha_m$. Then matching coefficients we get that $$\frac{f_j}{f_0}=(-1)^{j}\cdot\sigma_j.$$ Now in the polynomial ring ${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]$, the symmetric group $W=\mathfrak{S}_m$ acts by permuting variables, and a well known fact from invariant theory says that the invariant subring is generated by the elementary symmetric polynomials, i.e. $${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]^W={\mathbb{C}}[\sigma_1,\ldots,\sigma_m]={\mathbb{C}}\left[\frac{f_1}{f_0},\ldots,\frac{f_{m-1}}{f_0}\right].$$ Now the object $$\Delta^2=\prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)$$ is also invariant under $W$, hence we must have that $$\Delta^2=P\left(\frac{f_1}{f_{0}},\ldots,\frac{f_{m-1}}{f_0}\right)$$ for some polynomial $P\in{\mathbb{C}}[y_1,\ldots,y_m]$. Multiplying $\Delta^2$ by an appropriate power of $f_0$ will clear denominators to give you an honest polynomial in the $f_0,\ldots,f_m$. Certainly the exponent $m(m-1)$ will work, but it is not quite clear to me why $2(m-1)$ will, as you claim in your original statement...and perhaps this is the point of your question...?

Perhaps you will find the following helpful, which expands a bit on Robert Israel's answer: Write $f(x)=f_0x^m+\ldots+f_m=f_0\cdot(x-\alpha_1)\cdots(x-\alpha_m).$ Multiplying out the RHS we get $f(x)=f_0\cdot \sum_{j=0}^m(-1)^j\sigma_j(\alpha_1,\ldots,\alpha_m)\cdot x^{m-j}$ where $\sigma_j=\sigma_j(\alpha_1,\ldots,\alpha_m)$ is the $j^{th}$ elementary symmetric polynomial in $\alpha_1,\ldots,\alpha_m$. Then matching coefficients we get that $$\frac{f_j}{f_0}=(-1)^{j}\cdot\sigma_j.$$ Now in the polynomial ring ${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]$, the symmetric group $W=\mathfrak{S}_m$ acts by permuting variables, and a well known fact from invariant theory says that the invariant subring is generated by the elementary symmetric polynomials, i.e. $${\mathbb{C}}[\alpha_1,\ldots,\alpha_m]^W={\mathbb{C}}[\sigma_1,\ldots,\sigma_m]={\mathbb{C}}\left[\frac{f_1}{f_0},\ldots,\frac{f_{m-1}}{f_0}\right].$$ Now the discriminant $$\Delta^2=\prod_{1\leq i<j\leq m}(\alpha_i-\alpha_j)^2$$ is also invariant under $W$, hence we must have that $$\Delta^2=P\left(\frac{f_1}{f_{0}},\ldots,\frac{f_{m-1}}{f_0}\right)$$ for some polynomial $P\in{\mathbb{C}}[y_1,\ldots,y_m]$. Multiplying $\Delta^2$ by an appropriate power of $f_0$ will clear denominators to give you an honest polynomial in the $f_0,\ldots,f_m$. Certainly the exponent $m(m-1)$ will work, but it is not quite clear to me why $2(m-1)$ will, as you claim in your original statement...and perhaps this is the point of your question...?