We know that the path lifting theorem is true for covering maps. The map $p: \mathbb{R}_+ \to \mathbb{S}^1$ given by $p(t)=(\cos(2\pi t), \sin(2 \pi t))$ gives an example of a local homeomorphism which is not a covering map.
I want to know what really goes wrong in this case, due to which the path lifting theorem fails to hold good here. Thanks for any help.
Your example is the "standard" covering of $S^1$ by $\mathbb R$ restricted to $\mathbb R_+$. Since you state that this is a local homeomorphism, I assume that $0 \not\in\mathbb R_+$. From the parametrization, we see that $p$ wraps $\mathbb R_+$ around $S^1$ in counterclockwise orientation, starting at $1$. Now consider the path $$ \gamma:[0,1]\rightarrow S^1,\ t \mapsto ie^{-2\pi it}. $$ This path goes one time around $S^1$ in clockwise orientation, starting at $i$. This path can't be lifted through $p$ to the starting point $\frac{1}{4} \in \mathbb R_+$. Such a lift would move from $\frac{1}{4}$ downwards towards $0$ and "fall off" $\mathbb R_+$ at $0$ after a quarter of the way. So the path lifting theorem fails.
Here's a more detailed explanation of why a purported lift of $\gamma$ with starting point $\frac{1}{4}$ must move from $\frac{1}{4}$ straight down to $0$.
So let's assume that $\widetilde\gamma:[0,1]\rightarrow \mathbb R_+$ is a lift of $\gamma$ with starting pont $\frac{1}{4}$. This means we have $\widetilde\gamma(0) = \frac{1}{4}$, $p \circ \widetilde\gamma = \gamma$, and of course $\widetilde\gamma$ is continuous.
Denote by $T^1 = \{z \in S^1\ |\ Im\ z > 0 \}$ the open upper hemisphere and put $U := p^{-1}(T^1) \subseteq \mathbb R_+$. If we also put $J_k := (k,k+\frac{1}{2}) \subseteq \mathbb R_+$ for $k \in \mathbb N_0$, then we have $$ U = \bigcup_{k \in \mathbb N_0} J_k $$ and each of the intervals $J_k$ is a connected component of $U$. Moreover $p|_{J_k}:J_k \rightarrow T^1$ is a homeomorphism. Now, $\gamma([0,\frac{1}{4})) \subseteq T^1$, so we must have $\widetilde\gamma([0,\frac{1}{4})) \subseteq U$. Since $\widetilde\gamma([0,\frac{1}{4}))$ is connected and $\widetilde\gamma(0) = \frac{1}{4}$, we must have $\widetilde\gamma([0,\frac{1}{4})) \subseteq J_0$. So we have $$ \gamma|_{[0,\frac{1}{4})} = (p|_{J_0})\circ(\widetilde\gamma|_{[0,\frac{1}{4})}). $$ But $p|_{J_0}$ is a homeomorphism, so we have $$ \widetilde\gamma|_{[0,\frac{1}{4})} = (p|_{J_0})^{-1}\circ\gamma|_{[0,\frac{1}{4})}. $$ Now, as $t \in [0,\frac{1}{4})$ goes from $0$ to $\frac{1}{4}$, $\gamma|_{[0,\frac{1}{4})}(t)$ traces out the top right quarter of $S^1$ from $i$ to $1$. By looking at the construction of $p$, this implies that $(p|_{J_0})^{-1}\circ\gamma|_{[0,\frac{1}{4})}(t)$ traces out the line segment from $\frac{1}{4}$ down to $0$. So $\widetilde\gamma|_{[0,\frac{1}{4})}(t)$ approaches $0$ as $t$ tends to $\frac{1}{4}$ from below. So $\widetilde\gamma$ can't be continuously extended to $\frac{1}{4}$. Thus we have a contradiction and $\widetilde\gamma$ doesn't exist.