I would like to calculate $\mathrm{td}(X)$ for $X\in H^0(\mathbb P^n, \mathcal O(k))$. Since $$ 0\to TX\to T\mathbb P^n\to N\to 0, $$ we have $\mathrm{td}(X)=\mathrm{td}(\mathbb P^n)/\mathrm{td}(N)$ and we need to calculate $\mathrm{td}(N)$. Again, $$ 0\to -K_X\to -K_{\mathbb P^n}\to N\to 0 $$ and so $\mathrm{td}(N)\mathrm{td}(-K_X)=\mathrm{td}(-K_{\mathbb P^n})$, but $$ -K_X=\mathcal O(n+1-k)|_X,\ -K_{\mathbb P^{n}}=\mathcal O(n+1)|_X, $$ $$ N=\mathcal O(k)|_X $$ hence $\mathrm{td}(\mathcal O(a))\mathrm{td}(\mathcal O(b))=\mathrm{td}(\mathcal O(a+b))$ which is not true. (I use $\mathrm{td}(E|_X)=\mathrm{td}(E)|_X$ many times.)
Where did I make a mistake? What is the right way to calculate $\mathrm{td}(X)$?
The second exact sequence is not correct. After taking the $\mathrm{det}(\cdot)$ of each vector bundle in the first exact sequence we get $$ -K_X\otimes N=-K_{\mathbb P^n}|_X, $$
but not the second exact sequence. Hence there is no contradiction, we have $$ \mathrm{td}(N)=\mathrm{td}(\mathcal O(k))\ \mbox{and }\ \mathrm{td}(X)=\mathrm{td}(\mathbb P^n)/\mathrm{td}(\mathcal O(k)). $$