= Solution
For $v\in T_pS$, let $\gamma_v$ be the maximal <geodesic> satisfying
$$
\gamma_v(0)=p,
\qquad
\dot\gamma_v(0)=v.
$$
The <exponential map> is
$$
\exp_p(v)=\gamma_v(1)
$$
on the <domain of the exponential map>
$$
\mathcal D_p
=\{v\in T_pS:\gamma_v\text{ is defined throughout }[0,1]\}.
$$
This is an open, star-shaped neighbourhood of zero. In local coordinates the geodesic equation is a smooth ordinary differential equation, so smooth dependence on its initial position and velocity proves that $(p,v)\mapsto\exp_p(v)$ is smooth on its domain.
For both requested phenomena consider the embedded surface
$$
S=\{(x,y,0):(x,y)\ne(0,0)\}
$$
with $p=(1,0,0)$. Its geodesics are Euclidean straight lines for as long as they remain in $S$. The initial vector $v=(-2,0,0)$ would give
$$
\gamma_v(t)=(1-2t,0,0),
$$
which reaches the missing origin at $t=1/2$. Thus $v\notin\mathcal D_p$, and the domain is not all of $T_pS$.
The point $q=(-1,0,0)$ is not in the image of $\exp_p$. Any geodesic from $p$ to $q$ would have to be their unique Euclidean straight line, which passes through the omitted origin. Hence this exponential map is also not surjective, as summarized by <exponential map of the punctured plane>.
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