What is the essence of the constant factor in the standard definitions of the discriminant?
As Robert said, if you want everything to work in $\mathbb Z[f_0,\ldots,f_m]$, you need that factor. I'll also mention that your polynomial indexing is messed up, you probably meant the sum to go from $j=0$ to $j=m-1$, not $j=1$ to $j=m$.
In any case, things become clearer if you study the theory of resultants and ask when two polynomials have a common root. If they're not monic, you really want to view them as homogeneous polynomials that may have a common root "at infinity" if their leading coefficients both vanish. The discriminant of $f(x)$ is, essentially, the resultant of $f(x)$ and $f'(x)$.
The factor $f_0^{2m-2}$ makes the discriminant a polynomial in the coefficients $f_0, \ldots, f_m$.
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...?