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.
To apply the homotopy lifting property, you need to assume that $f: X\to B$ has a lifting. It doesn't assert that any homotopy can be lifted without assumptions. Your first bullet point is not correct.
However all 3 facts stated are correct (modulo some extra conditions on the topological spaces).
For a simple example highlighting why your argument doesn't hold, consider $f:\mathbb S^1\to\mathbb S^1$ the identity map, $\pi:\mathbb R\to \mathbb S^1$ the universal covering map. Then $f$ cannot be lifted, to $\tilde f:\mathbb S^1\to \mathbb R$, hence any homotopy between $f$ and another map cannot be lifted as well.
Best Answer
Path-connected won't work : otherwise take $Y=*$, then you would be claiming that any continuous function $T\to X$ lifts to $E$, which is known to be wrong.
However, if you require $T$ to be simply-connected and locally path-connected, then the answer will be yes, at least under the usual hypotheses of covering space theory.
Namely, here's a nice generalization of what you're looking for :
If $T$ is simply-connected, then the image of $\pi_1(Y\times T)$ will be the same as that of $Y$ and so your assumption implies that there is a unique lift on $Y\times T$ having the desired value on $(y,t_0)$. Note that composing with the inclusion at $t_0$, $Y\to Y\times T$ provides a lift of $f_{t_0}$, so it must be $\tilde f_{t_0}$ by uniqueness.