Let $M$ be an $n$-dimensional Riemannian manifold with sectional curvature lower bound 1. Fix a point say $O\in M$, let $S(r)$ denote the distance sphere centered at $O$ with radius $r$. The classical Hessian comparison theorem says that the principle curvatures of $S(r)$ is less than that of standard sphere ${S}^n(1)$. And Toponogov triangle comparison implies that given any two point in $S(r)$ there distance in $M$ is less than or equal to the correspond distance in round sphere with the same openning angle at $O$. 


So is there any way to see how the intrinsic diameter (i.e. the length metric induced from ambient metric) upper bound?


How about the Ricci curvature case?



Diameter estimate of distance sphere of positive curved manifold
I guess you want to ask is it true that 
$$\mathop{\rm IntrinsicDiameter}[S(r)]\le\mathop{\rm IntrinsicDiameter}[\tilde S(r)],$$
where $\tilde S(r)$ denotes the sphere of radius $r$ in the standard sphere.


* This is true if $r\ge \tfrac\pi2$; it follows since $S(r)$ has bigger curvature than $\tilde S(r)$ in the sense of Alexandrov.
* Note that if $r<\tfrac\pi2$ then $S(r)$ might be not connected; in this case 
$$\mathop{\rm IntrinsicDiameter}[S(r)]=\infty.$$ 
If sectional curvature $\ge 1$, I do not see other counterexamples.
It reminds me some questions related to the conjecture that boundary of Alexandrov space is an Alexandrov space.   
If you find a way to prove it then likely you will get some nontrivial corollaries of this conjecture say if $\Sigma$ is an Alexandrov space with curvature $\ge 1$ then $\mathop{\rm diam}\partial\Sigma\le \pi$ or perimeter of any triangle in $\partial\Sigma$ is at most $2{\cdot}\pi$.  

If $r\le\tfrac\pi2$, it is possible to construct a short map $h\_r\colon \tilde S(r)\to M$ so that its image covers $S(r)$.
In particular 
$$\text{area}[S(r)]\le\text{area}[\tilde S(r)]$$
(which is obvious anyway).
In general the image of $h\_r$ contains creases which stick inside $S(r)$ which in principle might be used as a shortcut.
* For Ricci curvature the statement does not hold even if $S(r)$ is connected.
You may take a small disc in hyperbolic plane and take a warp product with the sphere to make the Ricci curvature of obtained manifold to be colose to $+\infty$. The sphere $S(r)$ will have intrinsic diameter bigger than $\tilde S(r)$ as far as $S(r)\ne\emptyset$.


