$s(k)$ is made of $d^k$ numbers. They are labelled by the k-tuple $(j_1,j_2,\dots j_k)$ - and there are $d^k$ possibilities. 
For example, if $d$ is 3 and $k$ is 2, there are 9 values: $s^{(1,1)}$,  $s^{(3,2)}$ etc.

For a concrete example, the value of  $s^{(2,3)}$ is the integral $\int_{0<u_1<u_2<t}\,dx_2(u_2)dx_3(u_1)$ or $\int_0^t\int_0^{u_1}x_2'(u_2)x_3'(u_1)\,du_2\,du_1$ . I think you are using $x$ and $X$ for the same thing - the function from $[0,t]$ to $\mathbb{R}^d$ which defines the path. The numbers in the subscripts of $x$ in the integral determine which component of $s(k)$ is being calculated.

Let's say the path is piecewise linear between $N$ points, where $N>1$, and is specified as a $d \times N$ matrix $M$. $M$ is enough to calculate the signature - we don't need to know the exact speed the path is traversed, we don't need $t$ or $X$. Then let the signature of the straight path from the $i$th point to the $(i+1)$th point be $a_i$. For any $k$, and any $(j_1,\dots,j_k)$, we know that the value of $(a_i)^{(j_1,\dots,j_k)}$ is $\frac1{k!}\prod_{h=1}^k(M_{j_h,i+1}-M_{j_h,i})$ by explicitly doing the integrals. 

Let the signature of the whole of the path from the first point up to the $(i+1)$th point be $b_i$. We can calculate the value of $b_i$ "up to level $K$" (i.e. for all tuples $(j_1,\dots,j_k)$ with $k<K$) cumulatively in $i$, from the fact that $b_1=a_1$ and Chen's identity, which says that, for each $k$, and any $(j_1,\dots,j_k)$, $(b_{i+1})^{(j_1,\dots,j_k)}=\sum_{h=0}^k(b_i)^{(j_1,\dots\,j_h)}(a_{i+1})^{(j_{h+1},\dots,j_k)}$ . We then get the signature of the whole path,  $b_{N-1}$, up to level $K$.