The following is from Lecture notes of Professor Farrell. A hopefully working link is here(page 95):
https://www.dropbox.com/s/80n4wd6xctpe6yr/Characteristic%20Classes%20%28Sparkie%20%E7%9A%84%E5%86%B2%E7%AA%81%E5%89%AF%E6%9C%AC%202014-02-19%29.pdf
We now reach proposition 7. Let $E$ be a complex vector bundle, then the mod 2 reduction of the total Chern class of $E$ is the total Stiefel-Whitney class of $E$. Since there is no Chern classes in odd dimensions, we have there is no Stiefel-Whitney classes in odd dimension as well.
$\textbf{Proof}$
Suppose $E$ is a line bundle. Then we know $w_{0}(L)=1, w_{1}(L)=0,w_{2}(L)=e_{\mathbb{Z}_{2}}(L)=\phi(e(L))$. On the other hand we know $c_{0}(L)=1, c_{1}(L)=e(L)$. So this verified it for line bundles.
For sum of line bundles we have
$$
E=\oplus^{n}_{i=1}L_{i}
$$
So the total Chern class is
$$
c(E)=c(L_{1}) \cup \cdots \cup c(L_{n})\mapsto_{\phi}\omega(E)=\omega(L_{1}) \cup \cdots \cup \omega(L_{n})
$$
We are now going to use splitting principle. But there is a subtle point. We now discuss it.
By the splitting principle there exist
$$
f:\mathcal{B}\rightarrow B
$$
such that
$$
f^{*}(E)=\oplus_{i=1}^{n}L_{i}
$$ and
$$
f^{*}:H^{*}(B,R)\rightarrow H^{*}(\mathcal{B},R)
$$
is monic.
Therefore by naturality and the fact $f^{*}$ is monic with respect to $\mathbb{Z}_{2}$ coefficients.
$$
\phi(f^{*}(c(E)))=f^{*}(\phi(c(E)))=f^{*}(\omega(E))
$$
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 $c_2$.
Best Answer
Question: "Sorry but I have a couple of questions: Why is c(1)c(2)=c2(V) and why is c1(V)=c(1)+c(2)? And also, when we write c using these t's, they sort of serve to indicate what term corresponds to which ci right? @hm2020"
Answer: when $r=2$ the formula follows from your relation $c(V \otimes L)=\prod_j (1 + c_1(L_j') + c_1(L))$: Let $c(i):=c_1(L_i)$ and $c:=c_1(L)$. You get the calculation
$$c_t(V\otimes L)=(1+(c(1)+c)t)(1+(c(2)+c)t)= $$
$$...+(c(1)c(2)+(c(1)+c(2))c+c^2)t^2=$$
$$\cdots + (c_2(V)+c_1(V)c_1(L)+\binom{2}{2}c_1(L)^2)t^2=$$
$$c_0(V\otimes L) +c_1(V\otimes L)t+c_2(V\otimes L)t^2.$$
Note: You get
$$c_t(V)=(1+c(1)t)(1+c(2)t)= $$
$$1+(c(1)+c(2))t+c(1)c(2)t^2=c_0(V)+c_1(V)t+c_2(V)t^2$$
hence
$$c_1(V)=c_1(L_1)+c_1(L_2)\text{ and }c_2(V)=c_1(L_1)c_1(L_2).$$
Here you view the elements as "living" in a commutative ring and multiply (they all live in the even cohomology ring $H^{2*}(X)$ which is commutative) You find this explained in Hartshorne, Appendix A.
Note: Formulas for Chern classes $c_i(E\otimes F)$ where $E,F$ have "Chern roots" $a_i,b_j$ are expressed in terms of polynomials in the roots $a_i,b_j$. The coefficient of the $t^i$ term "lives" in the group $H^{2i}(X)$. Whenever you have a theory of Chern classes $c_i(E) \in H^{2i}(X)$ living in the even part of a cohomology ring $H^*(X)$ you may perform such calculations. The experession
$$c_t(E):=c_0(E)+c_1(E)t+\cdots + c_r(E)t^r$$
"lives" in
$$H^0(X)\oplus H^2(X)t\oplus \cdots \oplus H^{2r}(X)t^r.$$
For any rank $r$ locally trivial sheaf $E$ there is the complete flag bundle $F(E)$ of $E$ and an injection
$$H^*(X) \subseteq H^*(F(E)),$$
hence you may perform all calculations in $H^*(F(E))$. In $H^*(F(E))$ you have the equality
$$c_r(E)=c_r(L_1)+ \cdots + c_r(L_r)$$
where $L_i$ are invertible sheaves. You get the formula
$$c_t(E)=c_t(L_1)\cdots c_t(L_r)$$
in $H^{2*}(X)[t]$: It holds in $H^{2*}(F(E))[t]$ but since $c_t(E) \in H^{2*}(X)[t]$ the equality holds here.