Expected norm of first $d$ coordinates of the $n$-dimensional sphere

This is a classic example of linearity of expectation. Your random variable is $$X_d=frac{Y_1^2+cdots+Y_d^2}{Y_1^2+cdots+Y_n^2}$$ where the $Y_i$ are independent standard normal variables. That is $$X_d=Z_1+cdots+Z_d$$ where $$Z_i=frac{Y_i^2}{Y_1^2+cdots+Y_n^2}.$$ By symmetry, the expectations of the $Z_i$ are the same: $E(Z_1)=cdots=E(Z_n)=lambda$ say. But $$Z_1+cdots+Z_n=frac{Y_1^2+cdots+Y_n^2}{Y_1^2+cdots+Y_n^2}=1$$ so that $$E(Z_1)+cdots+E(Z_n)=E(Z_1+cdots+Z_n)=1,$$ and $lambda=1/n$. Then $$E(X_k)=E(Z_1+cdots+Z_k)=E(Z_1)+cdots+E(Z_k)=klambda=frac kn.$$

Call $A(d,n)$ this quantity. By symmetry, for any subset $I$ of size $d$ in ${1,dots,n}$, we have $E(|frac{(y_i)_{iin I}}{|Y|}|^2)= A(d,n)$. $$ sum_I E(|frac{(y_i)_{iin I}}{|Y|}|^2) = binom{n}{d} A(d,n),$$ where the sum is taken on all subsets $I$ of size $d$. But also, $$ sum_I E(|frac{(y_i)_{iin I}}{|Y|}|^2) =sum_I E(frac{1}{|Y|^2} sum_{iin I}|y_i|^2) = E(frac{1}{|Y|^2} sum_Isum_{iin I}|y_i|^2).$$ In the latter sum, the term $|y_1|^2$ appears exactly $binom{n-1}{d-1}$ times (to create a set $I$ containing $1$, it suffices to choose $d-1$ index in ${2,dots,n}$), and so does $|y_2|^2$, etc. Hence, we have $$ sum_I E(frac{1}{|Y|^2} sum_{iin I}|y_i|^2) = binom{n-1}{d-1}E(frac{1}{|Y|^2} sum_{i=1}^n|y_i|^2)= binom{n-1}{d-1}.$$ Hence, $binom{n}{d} A(d,n)= binom{n-1}{d-1}$ and $A(d,n)=frac d n$.