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 $c_2$.
Are you familiar with the Chern character? Its definition and basic properties are outlined here https://stacks.math.columbia.edu/tag/02UM.
Since we are on $\mathbb P^2$, I will write $c_1(E) = c_1H$ and $c_2(E) = c_2H^2$, where $c_1,c_2 \in \mathbb Z$ and $H$ is the hyperplane class. Now the Chern character of a rank $r$ bundle on $\mathbb P^2$ is
$$
r + c_1H + \frac{(c_1H)^2 - 2c_2H^2}{2} = r + c_1H + \frac{c_1^2 - 2c_2}{2}H^2.
$$
The reason we use the Chern character is because of nice formal properties like $ch(E\otimes F) = ch(E)ch(F)$. We have $ch(\mathcal O(n)) = 1 + nH$, so multiplying out (and using $H^3 = 0$) we get
$$
ch(E(n)) = r + (c_1 + rn)H + \frac{c_1^2 + 2nc_1 - 2c_2}{2}H^2.
$$
Now, writing $d_iH^i = c_i(E(n))$, we can immediately read off $d_1 = c_1 + rn$ (by comparing to the formula for $ch$ of an arbitrary bundle). Plugging this into the relation obtained from setting $d_1^2 - 2d_2$ equal to the coefficient of $H^2$, we get a relation for $d_2$ which, if my scribbled algebra is correct, can be solved to yield
$$
d_2 = c_2 + (r-1)nc_1 + \frac{r^2n^2}{2}.
$$
Best Answer
One reference would be Hirzebruch's Topological Methods in Algebraic Geometry. The way to work it out is with the splitting principle, assuming that $E$ splits as a sum of line bundles, $E=\oplus_{j=1}^r L_j$. If $c_1(L_j)=\delta_j$, then you write $c(E) = \prod_{j=1}^r (1+\delta_j)$. Then $c(E^*) = \prod_{k=1}^r (1-\delta_k)$ and $c(E\otimes E^*) = \prod_{j,k} (1+\delta_j-\delta_k)$.
Now you just have to write things out and count carefully. It might be easier to keep score if you do $c(E\otimes F) = \prod_{j,k} (1+\delta_j+\eta_k)$ and specialize later to $F=E^*$.