Let $p:(\tilde X, \tilde x_0)\rightarrow(X, x_0)$ be a covering map that is path-connected and locally path-connected. Is it true that given $\gamma$ and $\gamma'$, two continuous paths in $\tilde X$ starting and ending at $\tilde x_0$, if they are the same in $\pi_1(\tilde X, \tilde x_0)$, then the paths $p \circ \gamma$ and $p \circ \gamma'$ are the same in $\pi_1(X, x_0)$?
It should be false because if we consider the universal covering of $S^1$, $p:(\mathbb{R}, 0)\rightarrow(S^1, (1,0))$ with $p(t) = e^{2\pi i t}$, and take $\gamma(t) = 1_0(t)$, a constant path at $0$, and $\gamma'(t) = \begin{cases} (2t, & \text{if } t \in [0, 1/2]), \ (2(1-t), & \text{if } t \in [1/2,1]). \end{cases}$, we have $[\gamma(t)] = [\gamma'(t)]$ in $\pi_1(\mathbb{R}, 0)$. However, $p \circ \gamma(t) = 1_{(1,0)}(t)$ is a constant path at $(1,0)$, and $p \circ \gamma'(t)$ is a closed path at $(1,0)$ that completes two loops around $S^1$. Therefore, $p \circ \gamma(t)$ and $p \circ \gamma'(t)$ are not homotopic, and thus $[p \circ \gamma(t)] \neq [p \circ \gamma'(t)]$ in $\pi_1(S^1, (1,0))$.
To your example: The path $p\circ \gamma'$ walks around the circle once counter-clockwise, then once clockwise. Since for loops, we only fix the starting and ending point, you can easily imagine pushing back the "turning point" of this path in a counter-clockwise fashion, contracting $p\circ \gamma$.
In general: Any pointed continuous map $f\colon (X,x) \to (Y,y)$ induces a group homomorphism $$\pi_1(f)\colon \pi_1(X,x)\to \pi_1(Y,y), \quad [\gamma]\mapsto [f\circ \gamma].$$ Taking this for granted, since closed paths at $x$ in $X$ which are homotopic relative endpoints represent the same class in $\pi_1(X,x)$, their image must be the same.