Both of your questions can be answered by one simple fact that the elements of $p_*(\pi_1(\widetilde X, \widetilde x_0))$ precisely the classes of those loops in $(X, x_0)$ which lift to a loop in $(\widetilde X, \widetilde x_0)$. In fact this result is presented in Hatcher (the relevant part is in bold):
Proposition 1.31. The map $p_*: \pi_1(\widetilde X, \widetilde x_0) \to \pi_1(X, x_0)$ induced by a covering space $p : (\widetilde X, \widetilde x_0) \to (X, x_0)$ is injective. The image subgroup $p_*(\pi_1(\widetilde X, \widetilde x_0))$ in $\pi_1(X, x_0)$ consists of the homotopy classes of loops in $X$ based at $x_0$ whose lifts to $\widetilde X$ starting at $\widetilde x_0$ are loops.
The proof is given in Hatcher, but I present the proof of the part in bold for completeness.
Suppose $[\gamma] \in p_*(\pi_1(\widetilde X, \widetilde x_0))$, then $[\gamma] = p_*[\widetilde \gamma_1]$, for some loop $\widetilde \gamma_1$ in $(\widetilde X, \widetilde x_0)$. So, we have $\gamma \simeq p \circ \widetilde \gamma_1 = \gamma_1$ (say). $\gamma$ and $\gamma_1$ are homotopic as paths, so their liftings will also be homotopic as paths (by homotopy lifting property), that is, $\widetilde \gamma_1 \simeq \widetilde \gamma$. Now, since $\widetilde \gamma_1$ is a loop, so, is $\widetilde \gamma$.
Conversely, suppose $\gamma$ is a loop in $(X, x_0)$ that lifts to a loop $\widetilde \gamma$ in $(\widetilde X, \widetilde x_0)$, this means that $\gamma = p \circ \widetilde \gamma$ or $[\gamma] = p_*[\widetilde\gamma]$. So, $[\gamma] \in p_*(\pi_1(\widetilde X, \widetilde x_0))$.
Now, coming back to your questions, note that since $[h_0] \in p_*(\pi_1(\widetilde X, \widetilde x_0))$, $h_0$ lifts to loop.
That will not be enough : indeed the assumption on $\pi_1(Y,y_0)$ only tells us about the path-component of $y_0$, nothing else.
To get a specific counterexample, take $X\to B$ to be the $2$-sheeted connected covering of $S^1$ (so $z\mapsto z^2, S^1\to S^1$), $b_0 = 1, x_0 = 1$; and take $Y = S^1\sqcup \mathbb R$, $y_0 = 0 \in \mathbb R$ and $f: Y\to B$ defined by $id_{S^1}$ on $S^1$ and $x\mapsto \exp(ix)$ on $\mathbb R$
Then clearly, as $\mathbb R$ is contractible, the requirement for $\pi_1(Y,y_0)$ is satisfied, but there is no lift (a lift would yield a section of the $2$-sheeted covering, which does not exist)
So unless we require more than just information about the path-component of $y_0$, path-connectedness of $Y$ is necessary
Best Answer
It seems like they're talking about the other direction as trivial, and indeed it's fairly easy. Note since $f=p\circ f',$ functoriality gives $f_*=p_*\circ f'_*,$ and now $$f_*\pi_1(Y,y_0)=p_*f_*'\pi_1(Y,y_0)\subset p_*\pi_1 (X',x).$$