Right - I *think* I see now... you consider the two twisted loop algebras as subalgebras of $\mathfrak{sl}_n({\mathbb C}((t)))$, in fact as fixed points for the involutions $\sigma$, $\tau$ on this larger algebra.

Let $\alpha_0,\ldots ,\alpha_{2r-1}$ be the simple positive roots of $\tilde{A}_{2r-1}$ with $\alpha_0$ the affine node. The canonical central element $c$ is the sum of these simple co-roots. In the case of $\tau$, we can assume after conjugating that $e_{\pm\alpha_i}\mapsto e_{\pm\alpha_{2r-i}}$ where $1\leq i\leq 2r-1$, and then we are forced to choose $e_{\pm\alpha_0}\mapsto e_{\pm\alpha_0}$. Let $\beta_0,\ldots ,\beta_r$ be the simple co-roots in $\tilde{A}_{2r-1}^{(2)}$. We will use the involution $\sigma$ to identify the corresponding simple co-roots with certain elements of the untwisted affine Lie algebra. For $1\leq i\leq r-1$ we can identify $e_{\pm\beta_i}$ with $e_{\pm\alpha_i}+e_{\pm\alpha_{2r-i}}$; $e_{\pm\beta_r}$ identifies with $e_{\pm\alpha_r}$. (Note that these are all fixed by $\tau$.) For the affine root we identify $e_{\beta_0}$ with something like $[e_{\alpha_0},e_{\alpha_1}-e_{\alpha_{2r-1}}]$, and similarly for $e_{-\beta_0}$. (We need to obtain an element of the $(-1)$ eigenspace for $\tau$.) I don't think this is quite right but it does tell us what the $\beta_i^\vee$ are in terms $\alpha_0^\vee,\ldots ,\alpha_{2r-1}^\vee$: we have $$\beta_0^\vee = 2\alpha_0^\vee+\alpha_1^\vee+\alpha_{2r-1}^\vee,\; \beta_1^\vee=\alpha_1^\vee+\alpha_{2r-1}^\vee,\ldots ,\; \beta_{r-1}^\vee=\alpha_{r-1}^\vee+\alpha_{r+1}^\vee,\; \beta_r^\vee=\alpha_r^\vee.$$
Now we look at Kac's tables (or work it out directly) to see that the canonical central element is $\beta_0^\vee+\beta_1^\vee+2(\beta_2^\vee+\ldots +\beta_r^\vee)=2\sum_0^r \alpha_i^\vee$. So I obtain $2c$.

In the case of $\sigma$, we think of the affine roots $\beta_0,\ldots ,\beta_{r-1}$ as forming the type $D_r$ subsystem which gives us $\mathfrak{so}_{2r}\subset\mathfrak{sl}_{2r}$, and make $\beta_r$ behave as the affine node. After conjugating we can assume that $\sigma(e_{\pm\alpha_i})=-e_{\pm\gamma(\alpha_i)}$. Then $e_{\beta_{r-1}}$ identifies with $e_{\alpha_1}-e_{\alpha_{2r-1}}$ and similarly $e_{\beta_i}$ for $1\leq i\leq r-2$ identifies with $e_{\alpha_{r-i}}-e_{\alpha_{r+i}}$. To complete the type $D_r$ subsystem we identify $e_{\beta_0}$ with $e_{\alpha_{r-1}+\alpha_r}-e_{\alpha_r+\alpha_{r+1}}$. It only remains to observe that $e_{\alpha_0}\mapsto -e_{\alpha_0}$ and so the last root element is $e_{\beta_r}=e_{\alpha_0}$. Again this isn't quite right but it will suffice for determining co-roots:
$$\beta_0^\vee = \alpha_{r-1}^\vee+2\alpha_r^\vee+\alpha_{r+1}^\vee,\; \beta_1^\vee=\alpha_{r-1}^\vee+\alpha_{r+1}^\vee,\; \ldots ,\beta_{r-1}^\vee=\alpha_1^\vee+\alpha_{2r-1}^\vee, \beta_r^\vee=\alpha_0^\vee.$$
Once again we have $\beta_0^\vee+\beta_1^\vee+2(\beta_2^\vee+\ldots +\beta_r^\vee) = 2\sum_0^r\alpha_i^\vee=2c$.

So in both cases I obtain $2c$, but it is entirely possible that I have misunderstood your question and/or dualized somewhere (or got something wrong). But it seems to me that the central element you obtain in the twisted case will be something like the index times $c$.

EDIT: I see you are saying (as one might expect) that $\widehat{\mathfrak{sl}}_r$ is the untwisted affine Lie algebra. But then I really don't understand why you are taking the loop algebra over that. I must have a different edition of Kac's book to you - I don't have a Remark 8.6 and I only have a Thm. 8.5.

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