$\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$Take any $(x_1,x_2,\dots,x_N)\in(\S^n)^N$. 
Let $(-1,1)\ni t\mapsto X_1(t)\in\S^n$ be any smooth curve such that  $$X_1(0)=x_1.$$ For any $t\in(-1,1)$, let 
$$X(t):=(X_1(t),x_2,\dots,x_N)$$ and $$S(t):=f(X(t))=X_1(t)+x_2+\dots+x_N[\in\R^{n+1}],$$
so that $X(0)=(x_1,x_2,\dots,x_N)$ and 
$$S(0)=s:=x_1+x_2+\dots+x_N.$$ Let $v:=X'_1(0)$, so that $S'(0)=v$. Let 
$$S_1(t):=f_1(X(t))=e_1\cdot S(t),$$
where $\cdot$ denotes the dot product and $e_1$ is the first vector of the standard basis of $\R^{n+1}$. So, $S'_1(0)=e_1\cdot v$. So, for 
$$r(t):=\frac{f_1(X(t))}{\|f(X(t))\|}=\frac{S_1(t)}{\|S(t)\|}$$
we have 
$$r'(0)=\frac1{\|S(0)\|^2}\Big(\|S(0)\|S'_1(0)-S_1(0)\frac{S(0)}{\|S(0)\|}\cdot S'(0)\Big) \\ 
=\frac1{\|s\|^2}\Big(\|s\|e_1\cdot v-(e_1\cdot s)\frac{s}{\|s\|}\cdot v\Big).$$
Thus, the gradient in question is $g-(g\cdot x_1)x_1$, where 
$$ g:=\frac1{\|s\|^2}\Big(\|s\|e_1-(e_1\cdot s)\frac{s}{\|s\|}\Big).$$