Confusion about covering projection

This is quite a detailed question: I'm struggling to understand a few parts of a proof of the following Lemma. I've placed stars ($\bigstar$) where I'd like to draw your attention.

Lemma: Let $P:Y\to X $ be a covering projection and $ f: Z \to X $ a continuous map where $Z$ is a simply connected and locally path connected space. Suppose given base points $x_0, y_0, z_0$ of $X,Y,Z$ with $p(y_0) = x_0 = f(z_0) $. Then there is a unique continuous $ \tilde{f} : Z \to Y$ with $ p \tilde{f} = f$ and $ \tilde{f}(z_0) = y_0$.

The proof, summarised/paraphrased, is as follows:

Given $ z \in Z$, choose a path $u$ from $z_0$ to $z$ in $Z$. Let $ \tilde{u} $ be the unique lifting of $fu$ to a path in $Y$ starting at $y_0$, and define $ \tilde{f}(z) = \tilde{u}(1) $.

Can then show $ \tilde{f} $ is well-defined, and that it's the only possible mapping that could work. It remains to be shown that it is continuous.

To do so, let $ z \in Z$ and let $V$ be an open neighbourhood of $ \tilde{f}(z)$. Without loss of generality ($\bigstar$), we can assume $V$ is of the form $ h^{-1}(U \times {d}) $, where $U$ us an evenly covered neighbourhood of $ p\tilde{f}(z) = f(z)$ and $ h: p^{-1}(U) \to U \times D$ is a homeomorphism with $D$ discrete.

$f^{-1}(U)$ is an open neighbourhood of $z$, so contains a path connected neighbourhood $W$. For any $ z' \in W$, we can choose a path from $z_0$ to $z'$ of the form $u.v$ where $u$ is a path from $z_0$ to $z$ and $v$ takes values in $W$. Now $fv$ takes values in $U$, which is evenly covered by $p$. So its lifting to a path in $Y$ starting at $\tilde{f}(z)$ must take values in $V \ (\bigstar \bigstar)$, and in particular $ \tilde{f}(z') = \tilde{(u.v)}(1) \in V$. So $W \subseteq \tilde{f}^{-1}(V) $ and hence $\tilde{f}$ is continuous.

My confusions are as follows:

$ \bigstar $: Is the following reasoning as to why this does not lose generality correct? Given any $y = \tilde{f}(z)$, consider $ p(y) $. We can find an evenly covered neighbourhood $U$ containing $p(y)$. Then there is a homeomorphism $ h : p^{-1} (U) \to U \times D$ where $D$ is a discrete space. Then $ (p(y), d) \in U \times \{d\} $ for any $ d \in D$ and $ h^{-1} (p(y),d) \subseteq h^{-1} (U \times \{d\})$. But $ h^{-1}(p(y),d) = p^{-1} p(y) $ which contains $y$.

$ \bigstar \bigstar $: I'm confused with this bit. Intuitively, $ p^{-1}(U)$ 'looks like' lots of copies of $U$, and $V$ is one of those copies. But I can't turn this into a definitive reason why $ \tilde{v} $ (i.e. the lifting of $fv$) must be entirely contained within $V$. On a different note, am I correct in saying $ V \cong U $? I'm convinced there's a simple set-theoretic explanation, but it is eluding me.

Any help will be greatly appreciated - I understand that this is a long post.

Thank you.

The explanation you give for $\bigstar$ seems correct to me.

As for $\bigstar \bigstar$:
Since the trace of $\tilde v$ (that is the set of points $\tilde v ([0,1] )$ in $Y$ through which the path $\tilde v$ runs) is a connected set and since it is contained in the disjoint union of the open sets $h^{-1} (U\times\lbrace \delta \rbrace) \; (\delta \in D)$ it must be contained in exactly one of them, namely $V=h^{-1} (U\times\lbrace d \rbrace) $ since this is the one that contains $\tilde v(0)=\tilde f(z)$.

Finally, you are absolutely correct that $U$ and $V$ are homeomorphic.
The natural homeomorphism is the restriction $p|V:V\to U$, whose inverse is the continuous map $h^{-1}\circ j:U\to V $, where $j:U\to U\times D: u\mapsto (u,d)$ is the insertion of $U$ in the relevant slice of the product.

I've not had time to think about this for long, so I'd welcome any corrections to the following.

$\bigstar $: Yes, that's fine. Covering projections can seem a strange concept at first, but I find that it helps a lot to bear in mind the typical example $ p : \mathbb{R} \to S^1 $ with $ p(t) = e^{2 \pi i t} $, which I'm sure you'll have seen (imagine bending the real line into a helix, which projects down to cover the unit circle: our discrete space is just $ \mathbb{Z}$). Given any point on the helix, it's clear you can find an open interval around that point which is one of the 'copies' of the interval on $S^1$ that it projects down onto under $p$.

$ \bigstar \bigstar $: You have that $fv(t) \in U $ for all $t$. This means that $ (fv(t),d) \in U \times \{d\}$ for all $t$. Now, you should know that by the definition of 'evenly covered', you have that $ ph^{-1} (x,d) = x $ for all $x \in X$ and $d \in D$ (see here: Query about proof of homotopy lifting property). This means that $ ph^{-1}(fv(t),d) = fv(t) $ for all $t$. But, by uniqueness of lifting, this exactly gives us the lifting of $fv$: we must have that $ \tilde{v} = h^{-1}(fv(t),d) $. By the form of $V$, we have that $ \tilde{v}(t) \in V $ for all $t$.