I think this is a matter of convention because $(m+n)$ will be the highest spin irrep. present. What is certainly true, however, is that the sum is insufficient to specify all the information about how the particle transforms. For example the $(1,0)$ and $(1/2,1/2)$ representations are certainly not the same, though the sum matches.
It's still interesting, however, to ask about the relationship between these numbers and the Casimir you mention, which for reference is given by the square of the Pauli-Lubanski pseudovector, $W_\mu=\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}J^{\nu\rho}P^\sigma$ where $J^{\mu\nu}$ is the generator of the Lorentz group. The wiki article on this is actually fairly reasonable.
The most important point to note about $W_\mu$ is that it generates the little group, who's representations are what actually determine a particle's spin content. As described, for example, in the first volume of Weinberg's QFT series, the Wigner classification of particles essentially comes down to working out the representations of the little group. Massive and massless particles behave differently because the little group changes when the mass is zero, but I'll point out that the expression OP has quoted, $-m^2s(s+1)$ is for the case of a massive particle. Furthermore, we should keep in mind here that the $m$ in this expression is the mass of the particle, not the half-integer $m$ used in specifying the particle's Poincare representation...this $m$ comes out of squaring $W_\mu$, which produces something like $\sim P^\mu P_\mu \boldsymbol{J}^2$ so the square of the 4-momentum produces the particle's mass squared while the rest of the expression becomes the square of the generator of spatial rotations.
Now, to see the relationship between the sum of the pair of $SU(2)$'s and spin as defined by the little group, recall that the map between the generators $\boldsymbol A$ and $\boldsymbol B$ of the $SU(2)\times SU(2)$ and the Lorentz group generators, which I will now write in terms of the generators of boosts $\boldsymbol K$ and rotations $\boldsymbol J$, is given by $\boldsymbol A= \frac{1}{2}(\boldsymbol J+i\boldsymbol K)$ and $\boldsymbol B= \frac{1}{2}(\boldsymbol J-i\boldsymbol K)$.
The next statement I would like to make is: this implies $\boldsymbol J=\boldsymbol A+\boldsymbol B$ and hence $\boldsymbol J^2 = (\boldsymbol A+\boldsymbol B)^2$, and hence the spin content as defined by the little group and spin content as defined by this pair of $SU(2)$'s must be the same (the Casimir's are the same after all).
This statement is really the one that I wanted to make here since, as far as I'm aware, this is the precise relationship between the $SU(2)$ representations and the little group representations: the little group's Casimir matches that of the total $SU(2)\times SU(2)$ spin (up to a factor of the particles mass, and only in the case of massive particles, though I imagine a similar statement holds for massless particles).
The only lose end I think I've really left here is that I've written $(\boldsymbol A+\boldsymbol B)^2$, called it the total spin squared and then called it a day, glossing over the fact that the total spin operator should properly look like $(A\otimes 1+1\otimes B)$, as we know from the usual story of spin addition in quantum mechanics. The sum of the vectors $\boldsymbol A+\boldsymbol B$ doesn't quite look this way as written. Essentially, to see that it takes the usual form, we would need to play some games about changing how we write our indices. If you're studying SUSY as the OP indicates, this is similar to exchanging Lorentz indices for spinor indices. The details of this would, I think distract from the main point I have been trying to make though, so I will instead refer the reader back to chapter 5.6 in Weinberg's book.
