I made a number of relevant comments to this question on tigu's previous question.
Here they are again, lightly edited.
No, but this there is a true statement like this. When you describe $V_i$ as the locus where some sections are dependent, there are a lot of subtleties in that description. It is only literally correct if there are "enough" holomorphic sections and if those sections are "generic enough". Otherwise, you have to start talking about keeping tracks of multiplicities and signs.
The details of how you deal with the technicalities will be different depending on whether you are looking at algebraic/holomorphic sections (in this case, your best source is Fulton's "Intersection Theory") or whether you are looking at smooth sections (in this case, Milnor's "Characteristic Classes" has a good reputation, thought I haven't read it). I'll discuss a single example.
Look at $\mathbb{P}^2$ and let $U:= \mathcal{O}(1) \oplus \mathcal{O}(−1)$. The Chern class is $1−h^2$, where $h$ generates $H^2(\mathbb{P}^2, \mathbb{Z})$. The point is that $\mathcal{O}(−1)$ has no nonzero holomorphic sections. So all the holomorphic sections of $U$ are of the form $(0, \sigma)$ for $\sigma$ a section of $\mathcal{O}(1)$. In particular, any two holomorphic sections of $U$ are proportional everywhere on $\mathbb{P}^2$.
If we perturb the above sections so that they are not holomorphic then, I think, there will be are two curves $C$ and $C'$ on which they become dependent. One of these curve should be weighted positively and one negatively, and their degrees cancel in $H^2(\mathbb{P}^2)$. It would be fun to work out an example of this. UPDATE: I worked this out below and, at least for the two sections I tried, I got a single contractible sphere rather than two curves.
Similarly, if you look at one holomorphic section of $U$, it will vanish along an entire line in $\mathbb{P}^2$. If you perturb this section to be smooth, then it will vanish along a codimension $2$ subvariety, with multiplicity $−1$. You might want to read this question for a further understanding of how negative numbers show up as intersection multiplicities. That $-1$ explains that $-h^2$.
The true statement I allude to above should be something like "If $U$ has enough holomorphic sections, and $c_1(U)=0$, then $U$ is trivial." I do not know what the right definition of "enough" is; I would guess "globally generated".
Okay, let's actually work out the examples discussed above. We will describe a point of $\mathbb{P}^2$ using homogenous coordinates $(x:y:z)$. The fiber of $\mathcal{O}(-1)$ over $(x:y:z)$ is the line in $\mathbb{C}^3$ spanned by $(x,y,z)$. The fiber of $\mathcal{O}(1)$ over $(x:y:z)$ is the dual of that, in other words, $\mathrm{Hom}(\mathrm{Span}_{\mathbb{C}} (x,y,z), \mathbb{C})$. It is easy to give global holomorphic sections of $\mathcal{O}(1)$: Just take an element of $\mathrm{Hom}(\mathbb{C}^3, \mathbb{C})$ and restrict it to every fiber of $\mathcal{O}(-1)$. Let $\sigma_1$ be the section obtained by restricting $(x,y,z) \mapsto x$ and let $\sigma_2$ and $\sigma_3$ use the other two coordinate functions.
There are no holomorphic sections of $\mathcal{O}(-1)$, but there are lots of smooth sections. Let
$$\tau_1(x:y:z) = \frac{1}{|x|^2+|y|^2+|z|^2} ( x \overline{x}, y \overline{x}, z \overline{x}).$$
Define $\tau_2$ and $\tau_3$ similarly.
It turns out that $\sigma_1 \oplus \tau_1$ is not sufficiently generic is, but $\sigma_1 \oplus \tau_2$, so let's look at that. The section $\sigma_1$ vanishes whenever $x=0$, a line in $\mathbb{P}^2$. The section $\tau_2$ vanishes when $y=0$. However, if you check closely, you'll see that $\tau_2$ is anti-holomorphic, not holomorphic, in a neighborhood of its vanishing locus. So that vanishing should count negatively, and $\tau_2$ vanishes with multiplicity $-1$ on the line $y=0$. The section $(\sigma_1, \tau_2)$ vanishes at the point $(0:0:1)$, where $x=y=0$, and does so with sign $1 \times (-1) = -1$. This shows that $c_2 = -h^2$.
Similarly, let's look at the pair of sections $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$.
Assuming that $x$ and $y$ are nonzero, the ratio $\sigma_1/\sigma_2$ is $x/y$, while the ratio $\tau_3/\tau_2 = \overline{z}/\overline{y}$. So $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$ are proportional when $x/y = \overline{y}/\overline{z}$, in other words, when $x \overline{z} = |y|^2$. I didn't check carefully, but I believe that $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_1)$ are linearly dependent precisely when $x \overline{z} = |y|^2$, including in the degenerate cases where these coordinates vanish.
The set $\{ (x:y:z) \in \mathbb{P}^2 : x \overline{z} = |y|^2 \}$ is a sphere. I can parameterize it as $(e^{i \theta} (1+\sin \phi) : \cos \phi : e^{i \theta} (1-\sin \phi))$ where $\theta$ is the longitude coordinate, running through $\mathbb{R}/(2 \pi \mathbb{Z})$, and $\phi$ is the lattitude coordinate, living in $[-\pi/2, \pi/2]$. I claim that this sphere is contractible in $\mathbb{P}^2$. To see this, map $S^2 \times [0,1]$ to $\mathbb{P}^2$ by $(\theta, \phi, t) \mapsto (e^{i \theta} (1+\sin \phi) : (1-t) \cos \phi : e^{i \theta} (1-\sin \phi))$. At one end of the homotopy, we have the previously described embedding. At the other end, we have the line segment $\{ (1+r:0:1-r) : r \in [-1, 1] \}$. This line segment is obviously contractible.
So the space where $(\sigma_1, \tau_2)$ and $(\sigma_2, \tau_3)$ are dependent has homology class $0$, showing that $c_1=0$.
1) For a line bundle $E$ the notation $E^{-1}$ is used for the dual $E^*$ because of the canonical isomorphism of line bundles $$ E\otimes E^*\stackrel {\cong }{\to } X\times \mathbb C: x\otimes \phi \mapsto \phi(x) $$
This says that in the Picard group $Pic(X)$ of $X$ , which is the group of equivalence classes of holomorphic line bundles on $X$, (the class of) $E^*$ is the inverse $E^{-1}$ of $E$ in the group-theoretic sense.
2) As Adam wrote in the comment, the equality of integers $h^1(E)=h^0(E^*\otimes K)$ comes from Serre duality.
Serre duality more precisely says that there is a non-degenerate pairing
$$ \Gamma(X,E^*\otimes K)\otimes H^1(X,E)\to \mathbb C $$
from which follows that each of the two finite dimensional $\mathbb C$-vector spaces $\Gamma(X,E^*\otimes K)$ and $ H^1(X,E)$ can be considered the dual of the other and this of course implies $h^1(E)=h^0(E^*\otimes K)$
3) Here are a few example of use of Riemann-Roch coupled with Serre duality.
a) If $E=\mathcal O$, you get $h^1(\mathcal O)=h^0( K)$: these are the two possible definitions of the genus $g$ of $X$ .
b) From $h^0( K)-h^1( K)=1-g+deg (K)$ you get $g-h^0( K^*\otimes K)=g-1=1-g+deg (K)$ so that $deg(K)=2g-2$.
This is a very concrete result: it says that if you take any non-zero meromorphic one-form $\omega$ its divisor (of zeros and poles) $div(\omega)$ will have degree $2g-2$ .
c) The best thing a cohomolgy group can do to please us mathematicians is to vanish!
Serre duality implies that a line bundle $E$ of degree $deg(E)\geq 2g-1$ will have $h^1(E)=h^0(E^*\otimes K)=0$ . Indeed the line bundle $E^*\otimes K$ only has zero as holomorphic section since its degree is negative: $-deg (E)+2g-2\lt 0$.
Riemann-Roch then gives the completely explicit formula involving no cohomology: $$h^0(E)=1-g+deg (E)$$
Best Answer
As Asal Beag Dubh says in the comments, the key point is to use the splitting principle to reduce the computations to the case of line bundles. Everything becomes more or less an exercise in symmetric function theory. Here are the four basic examples:
The dual bundle
If $L$ is a line bundle with Chern class $c_1(L)$, then the dual line bundle $L^{\ast}$ is isomorphic to the inverse line bundle and hence has Chern class $c_1(L^{\ast}) = - c_1(L)$. If $V \cong \bigoplus_i L_i$ then $V^{\ast} \cong \bigoplus_i L_i^{\ast}$, so we have
$$c(V^{\ast}) = \prod_i (1 - c_1(L_i))$$
from which it follows that
$$c_i(V^{\ast}) = (-1)^i c_i(V).$$
The tensor bundle
If $L, L'$ are line bundles with Chern classes $c_1(L), c_1(L')$, then the tensor product $L \otimes L'$ has Chern class $c_1(L \otimes L') = c_1(L) + c_1(L')$. If $V \cong \bigoplus_i L_i$ and $V' \cong \bigoplus_j L_j'$, then
$$V \otimes V' \cong \bigoplus_{i, j} L_i \otimes L_j'$$
so we have
$$c(V \otimes V') = \prod_{i, j} (1 + c_1(L_i) + c_1(L_j')).$$
Extracting more explicit formulas from this is a tedious exercise. An alternative here is to use the Chern character, which is multiplicative with respect to tensor product by design:
$$\text{ch}(V \otimes V') = \text{ch}(V)\text{ch}(V').$$
For example, this gives
$$c_1(V \otimes V') = c_1(V) \dim V' + c_1(V') \dim V.$$
You can get corresponding formulas for the hom bundle using the isomorphism $V^{\ast} \otimes W \cong \text{Hom}(V, W)$.
The symmetric powers
It's cleanest to do all of the symmetric powers at once. The key is the isomorphism
$$S(V \oplus W) \cong S(V) \otimes S(W)$$
where $S(V) \cong \bigoplus_i S^i(V)$ is the symmetric algebra. This is an isomorphism of graded vector bundles, and remembering the grading is important in what comes next. If $V \cong \bigoplus_i L_i$, it follows that
$$S(V) \cong \bigotimes_i S(L_i)$$
and hence that the graded Chern character of $S(V)$, as a graded vector bundle, can be computed as
$$\text{ch}(S(V)) = \sum_k t^k \text{ch}(S^k(V)) = \prod_i \text{ch}(S(L_i)) = \prod_i \frac{1}{1 - t e^{c_1(L_i)}}$$
where $t$ is a formal variable. Again, extracting more explicit formulas from this is a tedious exercise.
The exterior powers
As for the symmetric powers, we again have
$$\Lambda(V \oplus W) \cong \Lambda(V) \otimes \Lambda(W)$$
where $\Lambda(V) \cong \bigoplus_i \Lambda^i(V)$ is the exterior algebra. The discussion is exactly the same as for the symmetric algebra except that the last Chern character computation is a bit different, and we get
$$\text{ch}(\Lambda(V)) = \sum_k t^k \text{ch}(\Lambda^k(V)) = \prod_i \text{ch}(\Lambda(L_i)) = \prod_i (1 + t e^{c_1(L_i)}).$$
Once again, extracting more explicit formulas from this is a tedious exercise. To get you started on $c_2(\Lambda^2(V))$, by looking at the coefficient of $t^2$ we get
$$\text{ch}(\Lambda^2(V)) = \sum_{i < j} e^{c_1(L_i) + c_1(L_j)}.$$
The first term of this expansion gives you the dimension of $\Lambda^2(V)$, which you already know. The second term gives you the first Chern class, which is
$$c_1(\Lambda^2(V)) = (\dim V - 1) c_1(V).$$
The third term gives you the third term in the Chern character of $\Lambda^2(V)$, which you need to correct a little by $c_1^2$ to get $c_2$.