Yes, that is a fine proof.
As a general principle, if you can write down an isomorphism of vector spaces without making any choices, then you have written down an isomorphism of vector bundles.
Let's make that principle precise in this case. On a small open set $U$, trivialize the vector bundles $V_1$ and $V_2$ with frames (:=sections which form a basis at every point-- note that, seemingly contra the principle, we've made a choice) $e_1, \ldots, e_n$ and $f_1, \ldots, f_m$ (I changed your notation $r_1, r_2$ to $m, n$ to avoid double subscripts).
Then the bundle $V_1 \otimes V_2$ is trivialized with frame $e_i \otimes f_j$ and its determinant trivialized with the singleton frame (alphabetical order)
$$
(e_1 \otimes f_1) \wedge (e_1 \otimes f_2) \wedge \ldots \wedge (e_1 \otimes f_m) \wedge(e_2\otimes f_1) \wedge \ldots \wedge (e_n \otimes f_m)
$$
which we may map to
$$
(e_1\wedge \ldots \wedge e_n)^m \otimes (f_1 \wedge \ldots f_m)^n.
$$
Now it seems so far that our map depends on choices, but it does not. We just need to check that multiplying an $e_i$ by an invertible function, adding $\phi e_i$ to $e_j$ (for $\phi$ an arbitrary function), or flip-flopping $e_i$ and $e_j$ does nothing (and same for $f_i$ and $f_j$). Note that in each case, the given bases for $\text{det}(V_1 \otimes V_2)$ and $\text{det} (V_1)^m \otimes \text{det} (V_2)^n$ multiply by the same scalar function, so the map doesn't change.
We were (allegedly) doing all the above reasoning on a small open set $U$ -- otherwise, there may not exist a frame (the existence of a frame on an open set being equivalent to a vector bundle being trivial). Now suppose we define a global map by doing the same reasoning on ALL open sets. We have to check that if $U$ and $W$ are different, we have defined the same map on $U \cap W$.
But it follows from the independence of choices. The restriction of a choice of frame over $U$ to a frame over $U \cap W$ gives the map of bundles over $U \cap W$, and so does the restriction of a choice of frame over $W$. But we know that the map doesn't depend on a choice. So we've therefore given a global map of vector bundles.
In general it can be general to try to work even more abstractly, i.e. not in terms of picking bases, so that it becomes completely automatic, by the above principle, that a map of vector spaces extends to a map of vector bundles.
The nice relation that you hope for is not true unfortunately. Here is a counter-example: consider the trivial rank 2 vector bundle $E= \mathcal{O} \oplus \mathcal{O}$ over $\mathbb{CP}^{1}$. Then $$\mathbb{P}(E) \cong \mathbb{CP}^{1} \times \mathbb{CP}^{1}.$$
Now, note that $c_{2}(E)=0$ since $E$ is a trivial bundle, but $\int_{\mathbb{P}(E)} c_{2}(T\mathbb{P}(E)) = 4$, since this is equal to the topological Euler characteristic of $\mathbb{P}(E)$ (it is a standard result that the integral of the top Chern class of a complex manifold is equal to the topological Euler characteristic).
In general, I don't think that there will be a nice relation (although I could be wrong) since when $E$ has different Chern numbers, the topological type of $\mathbb{P}(E)$ will be different in general, so we are not even talking about Chern classes on the same complex manifold. However, you can actually compute some topological invariants of $\mathbb{P}(E)$ from the chern classes of $E$: see Proposition 15 of Okonek and Van de Ven's "Cubic forms and complex 3-folds".
Best Answer
1) The answer to your first question is that by definition the rank $r$ vector bundle $E$ on $X$ is ample if the line bundle $\mathcal O_{\mathbb P(E)}(1)$ on $\mathbb P(E)$ is ample.
[ If $ r=1$ we apparently have two definitions for the amplitude of the line bundle $E$, but they clearly coincide since $\mathbb P(E)=X$ and $O_{\mathbb P(E)}(1)=E$.]
2) An important property is that a quotient bundle of an ample bundle is ample.
This allows to negatively answer your second question, whether a bundle whose determinant is ample is ample:
Take $X=\mathbb P^1$ and $E=\mathcal O(-1)\oplus \mathcal O(2)$.
Then $\det(E)=\mathcal O(1)$, which is an ample line bundle.
Nevertheless $E$ is not ample because one of its quotients is $\mathcal O(-1)$, which is certainly not ample.
[In the other direction it is true, however, that the determinant of an ample bundle is ample.]
3) Another important property is that the restriction $E|Y$ of an ample vector bundle on $X$ to any subvariety $Y\subset X$ is ample.
This immediately implies that for a vector bundle $E$ of rank $\gt 1$ the lifted bundle $f^*E$ on $\mathbb P(E)$ is never ample since its restriction to a fiber of $f$ is trivial, and thus certainly not ample.
This answers your third and last question, again negatively.
Edit
Although ample line bundles are an older concept, inextricably linked to the venerable notion of divisors, ample vector bundles of rank $\gt 1$ were introduced by Hartshorne only in 1966, in this article.
He gives another definition of "ample", but the one I chose (because it is closer to the notions in Jim's question) is equivalent by Proposition 3.2 of the article.
By the way, ample vector bundles of rank $\gt 1$ are not mentioned in Hartshorne's celebrated book and as an ironic consequence his generalized concept of "ample" is not as well known as it deserves...