(Let's start with the complex case.)
Basically there are two issues here.
1) First of all, Chern classes are stably invariant, i.e. can't distinguish stably isomorphic vector bundles. So the question becomes
Do Chern classes determine an element in the group $\tilde K^0(X)$ of vector bundles on $X$ modulo stable equivalence?
(Example. Since $\pi_4(U(2))=\pi_4(S^1\times S^3)=\mathbb Z/2$ there is a non-trivial 2-bundle on $S^5$. But of course $\tilde K^0(S^5)=0$ so this bundle is stably trivial — in particular, it has zero Chern classes.)
2) This 'stable' variant of the question is (still very nontrivial but) much more tractable (essentially because K-theory is a generalized cohomology theory etc). The Chern character gives an isomorphism $K^0(X)\otimes\mathbb Q\to H^{\text{even}}(X;\mathbb Q)$. So
if $K(X)$ has no torsion, an element of $K(X)$ is completely determined by its Chern classes.
And if $H(X)$ is torsion-free so is $K(X)$ (this follows from AHSS). That's true e.g. for complex Grassmannians and flag manifolds.
(Example. The if $L\in\tilde K^0(\mathbb RP^6)$ is the complexified tautological bundle, $4L$ is a non-trivial element, i.e. a stably non-trivial bundle, with trivial Chern classes.)
For the first question I think you have some misunderstandings. Firstly Chern classes are defined for complex vector bundles - or equivalently principal $U(n)$-bundles. If $G$ is an arbitrary compact Lie group then the Chern classes of a principal $G$-bundle will be understood in terms of some representation $\rho:G\rightarrow U(n)$ (these always exist). For instance when $G=SU(n)$ the representation $\rho$ is the canonical inclusion.
Now if you are given a principal $G$-bundle $p:P\rightarrow X$ over a suitable space $X$ for some topological group $G$, and a representation $\rho:G\rightarrow U(n)$ then you can form the associated vector bundle $P\times_\rho\mathbb{C}^n\rightarrow X$ with total space
$P\times_\rho\mathbb{C}^n=P\times \mathbb{C}^n/[(e,v)\sim(eg,\rho(g)^{-1}v)]$
where $e\in P$, $v\in\mathbb{C}^n$ and $g\in G$. You can now ask for characteristic classes of this vector bundle, and in particular its Chern classes (and there are an infinte number of Chern classes, not just two).
Note that the Chern classes of this bundle are not the characteristic classes of the principal bundle $P$, but are the characteristic classes of some principal $U(n)$-bundle $Q\rightarrow X$ that admits a $(G,\rho)$-reduction of structure to $P$.
The principal bundle $P$ will have its own set of characteristic classes deriving from the cohomology of the classifying space $BG$ of $G$.
Note in particular that the vector bundle $P\times_\rho\mathbb{C}^n$ is not the adjoint bundle $ad(P)$, which is another vector bundle whose characteristic classes have even less to do with those of $P$. Consider for instance the canonical line bundle $\eta$ over $\mathbb{C}P^1=S^2$. Since $U(1)$ is abelian we have that $ad(\eta)\cong S^2\times \mathbb{R}$ is trivial (recall that the Lie algebra of $U(1)$ is $i\mathbb{R}$). However $\eta$ itself is not trivial and is in fact associated to the Hopf $S^1$-bundle $S^3\rightarrow S^2$ which has first Chern class $1\in H^2(S^2)$.
In summary it makes equal sense to ask for the characteristic classes of a principal bundle or the characteristic classes of a vector bundle with structure group $G$ if the representation $\rho$ is understood.
As for the second question, instantons cannot have a Chern class. Instantons are generally understood as certain field configurations. Given for instance as a suitable connection on a principal of vector bundle. I think what you are referring to is the instanton charge, which in the case above is defined to be the Chern class of the bundle to which the instanton is related. As I understand it a $U(1)$-instanton can not have non-zero second Chern class. It can have non-zero first Chern class. In fact for rank $N=1$ we have $c_1=n$ and $c_2=0$.
*Edit: Added to clarify comments.
I wasn't quite clear above. In the following I'll limit attention of $U(n)$-bundles. There is a one-to-one correspondence between (complex) vector bundles over $X$ and $U(n)$-bundles over $X$. This happens in one direction through the construction I gave above, and in the other by taking the frame bundle of a given vector bundle.
Thus, given a complex vector bundle $E\rightarrow X$ of rank $n$ we have a principal $U(n)$-bundle $Fr(E)\rightarrow X$ and an isomorphism of vector bundles $Fr(E)\times_{U(n)}\mathbb{C}^n\cong E$ over $X$. We can then take a connection on either bundle and ask for the Chern classes it gives. To see that the classes defined by each are the same recall firstly that a connection on a principal bundle induces one on any associated bundle. So a connection $A$ on the principal bundle $Fr(E)$ induces one, $\nabla_A$, on the vector bundle $Fr(E)\times_{U(n)}\mathbb{C}^n\cong E$. Moreover it is seen that both connections $A$ and $\nabla_A$ yield identical curvature 2-forms. Thus
$c_i(E)^{\nabla_A}=c_i(Fr(E))^A\in H^*(X)$
where we add a superscript to denote which connection we use to define the classes.
Now recall that the Chern classes are independent of the choice of connection. It follows that any other connection $A'$ on $Fr(E)$ yields identical Chern classes
$c_i(Fr(E))^A=c_i(Fr(E))^{A'}\in H^*(X)$.
By the same reasoning any other connection $\nabla'$ on your original vector bundle $E$ yields identical Chern classes
$c_i(E)^{\nabla'}=c_i(E)^{\nabla_A}\in H^*(X)$.
The point is that it doesn't matter which bundle $E$ or $Fr(E)$ we use to define the Chern classes: they are the same. The one-to-one correspondence between vector bundles and principal bundles carries over to their characteristic classes and you can consider the Chern classes to 'belong' to either bundle.
As for your comments regarding the Yang-Mills functional. Firstly I believe what you say about it being bounded by the second Chern class is only relevant to Yang-Mills theory over a 4-dimensional base. It is certainly not true for Yang-Mills on a surface, or 8-manifold, say. As for what bundle, it is the one you are working with, which should be granted by deeper context. By the above, it does not matter it is a vector bundle or the corresponding principal bundle.
And if you are talking about some $G$-bundle then yes, there must be an implicit unitary representation. The Chern classes are the characteristic classes belonging to $U(n)$. They can be defined as certain cohomology classes in the cohomology of the classifying space $BU(n)$. You cannot ask for the Chern classes of a $G_2$-bundle, say, which will have its own set of characteristic classes. If you are working with a vector bundle which has a $G$-structure then this $G$-structure provides a unitary representation for $G$.
*Edit 2: Clarification of further comments.
The Chern classes may be defined as certain classes in the cohomology of the classifying space $BU(n)$ of $U(n)$. The integral cohomology of $BU(n)$ is the polynomial ring $H^*(BU(n))\cong \mathbb{Z}[c_1,\dots c_n]$ on classes $c_i\in H^{2i}(BU(n))$. The classes $c_i$ are the universal Chern classes: the Chern classes belonging to the universal $U(n)$-bundle $EU(n)\rightarrow BU(n)$ with contractible total space (you can think of $BU(n)$ as the Grassmanian of $n$-planes in $\mathbb{C}^\infty$).
Then if $f:X\rightarrow BU(n)$ is a suitable classifying map for the bundle $E\rightarrow X$ (principal $U(n)$ or complex vector bundle, we have already decided these are the same things in a certain sense), in that $f^*EU(n)\cong E$, you can define the Chern classes of $E$ to be
$c_i(E)=f^*(c_i)\in H^{2i}(X)$.
This is the point of view in algebraic topology, and its truly quite marvelous that this same information can be extracted by studying the geometric data contained in a connection on $E$.
It may have been misleading when I said that there were an infinte number of Chern classes. On the one hand it's true if you consider the infinite unitary group which has $H^*(BU(\infty))\cong\mathbb{Z}[c_1,\dots,c_n,\dots]$, and you can certainly embed a rank $n$ vector bundle $E$ into the rank $n+1$ vector bundle $E\oplus\epsilon^1$ and ask for its Chern classes. However whilst you can take $n$ all the way to infinity in this manner, there will only be at most $n$ non-trivial Chern classes for a rank $n$ bundle. There are an infinite number you can ask for, but only $n$ interesting ones.
Now more generally you may be interested not in the individual Chern classes but in certain polynomial combinations of them, and there are indeed an infinite number of these you could potentially ask for.
Now the final point is that over a 4-manifold $X$ there will only be two potentially non-trivial Chern classes. This is simply because $H^*(X)$ vanishes in dimensions greater than $4$. Thus we have $c_1(E)\in H^2(X)$ and $c_2(E)\in H^4(X)$ and anything else lives in higher degrees and is thus trivial.
An application of this? If $E\rightarrow X$ is a rank $n$ complex vector bundle over a $4$-manifold with $n>2$, then there is a rank $2$ vector bundle $E'\rightarrow X$ and a bundle morphism $E\cong E'\oplus\epsilon^{n-2}$.
Best Answer
The answer to your first question is no, you cannot compute characteristic classes from the weights. In what follows I assume you are familiar with a little bit of complex/algebraic geometry.
Let $C$ be a Riemann surface of genus $g$, and $L$ any complex line bundle over $C$. (Topologically there will be one complex line bundle for every element in $H^{2}(C,\mathbb{Z}) \cong \mathbb{Z}$.)
Consider $X = \mathbb{P}_{C}(L \oplus \mathcal{O})$ (i.e. the projectivisation of the rank 2 complex vector bundle $L \oplus \mathcal{O}$ over $C$).
There is a natural Hamiltonian $S^{1}$-action on this space given by $z.[v_{1},v_{2}] = [v_{1},zv_{2}]$ (which is well-defined because the second factor is trivial). Furthermore the subset $S$ where $v_{2}=0$ is isomorphic to $C$, and consists of fixed points for the $S^{1}$-action. Furthermore the action has weight $1$ along $S$ (since the action is semi-free). One can see that the normal bundle of $S$ in $X$ is isomorphic to $L$. Since the degree of $L$ can take any value here, this example shows that degree of the normal bundle definitely cannot be recovered from the weight of the action.
A summary of what is true (at least for a fixed submanifold for Hamiltonian $S^{1}$-actions) is:
The class of the normal bundle plus the weights define a neighbourhood of the fixed submanifold up to equivariant symplectomorphism.
There are certain global localisation theorems, almost all coming from the Atiyah-Bott Localisation theorem, which state that the sum of certain combinations of weights and characteristic classes of the normal bundles of fixed submanifolds, summed over ALL of the fixed point data are equal to certain characteristic classes of the original manifold.