The map $f:T_1M\to \hat{\mathbb{R}}$ is continuous. Let $(M,d)$ be a complete Riemannian manifold and let $T_1M=\{v\in TM: \|v\|=1\}$. Define a map, $$s:T_1M\to \hat{\mathbb{R}},~~ s(v)= \sup\{t:d(\pi(v),\operatorname{exp}(tv))=t\}$$ where $\pi$ is the  projection map from the the tangent bundle to $M$ $(\pi(p,v)=p,~(p,v)\in TM)$. I need to prove that the map $s$ is continuous and $\hat{\mathbb{R}}$ is the one point compactification of $\mathbb{R}$.
Any reference or help will be appreciated.  

This is an extended comment.
I would call your function $s$ the cut distance function on the unit sphere bundle $T_1M$.
The claim is not true without further assumptions.
Consider, for example, a banana-shaped open set $M\subset\mathbb R^2$ with the Euclidean metric.
When the geodesic starting at $v$ is tangent to a concave part of the boundary, problems arise.
If you keep the base point fixed and approach the direction $v$ from different directions, you get different limits.
The basic problem is that the exponential map is not defined for all time.
This example and your definition of the function $s$ seems to suggest the assumption that all geodesics are defined for all time.
In other words, $M$ is assumed (geodesically or metrically; cf. Hopf–Rinow) complete.
If it is true, perhaps something like this would work:
Suppose $s$ was discontinuous at a point $v\in T_1M$ where $s(v)<\infty$.
Then there is a sequence of points $v_k\in T_1M$ and $\epsilon>0$ so that $v_k\to v$ and $s(v_k)\geq s(v)+\epsilon$ for all $k$ or $s(v_k)\leq s(v)-\epsilon$ for all $k$.
(To see this, recall that continuity means that if $v_k\to v$, then $s(v_k)\to s(v)$. If this fails, there is a sequence so that the images do not converge to the right limit or at all; then take a suitable subsequence.)
