What is the meaning of X is of order 2N in blind rotation of TFHE?

Everything they are saying is that you are always performing your polynomial operations modulo $X^N + 1$.

In this case, you have $X^N \equiv -1 \pmod {X^N+1}$ and therefore,

$X^{2N} \equiv (X^N)^2 \equiv (-1)^2 \equiv 1 \pmod {X^N+1}$.

Moreover, you can see that for any $0 < k < 2N$, $X^{k} \not \equiv 1 \pmod {X^N+1}$. That is, $2N$ is the smallest positive integer that maps $X$ to $1$ when used as the exponent. That is what we call the (multiplicative) order of an element.

Now, when they say that those sums are performed modulo $2N$, that is because the sum is computed "on the exponent" of $X$. That is, to compute $\sum_{j = 1}^{n}\bar{s}_j \bar{a}_j$, they actually compute $$ \prod_{j = 1}^{n} X^{\bar{s}_j\bar{a}_j} = X^{\sum_{j = 1}^{n}\bar{s}_j \bar{a}_j} $$

But because the order of $X$ is $2N$, what you really get is $$ X^{\sum_{j = 1}^{n}\bar{s}_j \bar{a}_j \bmod{2N}} $$