Covariant derivative in associated bundle Let $\pi:P→M$ be an arbitrary principal $G$-bundle, $F$ a vector space  and $\pi_F : P × _G F → M$ the associated fiber bundle. 
If $U$ is an open subset of $M$, then a smooth map
$ϕ : \pi^{−1} (U)→ F$ is said to be equivariant  if
$ϕ(p · g) = g^{-1} · ϕ(p)$
Given a section $s : V → \pi_F^{-1}(U) $  we can show that there is a  one-to-one and onto correspondence between $ϕ(p)$ and $s$.
Assuming now that $\pi:P→M$ has defined on it a connection form $ω$, we define the covariant exterior derivative 
$d^ω ϕ$ of $ϕ$ by
$(d^ω ϕ)_p(v)= (dϕ)_p(v^H)$
Where $dϕ$ is the exterior derivative of $ϕ$ , $v ∈ T_pP$ and $v^H$ is the horizontal part of it.
It also can be show that 
$d^ω ϕ=dϕ+ω\cdotϕ$ where $ω\cdotϕ$ means $ω$ acting on $ϕ$
Now given a section on the principal bundle $\sigma : V → \pi^{-1}(U)$
we can show also that   
$(\sigma^*d^ωϕ)(T)=d(\sigma^*ϕ)(T)+(\sigma^*ω)(T)\cdot\sigma^*ω$ 
Identifying  $(\sigma^*d^ωϕ)(T)$ as $\nabla_T S $ where $S=\sigma^*ϕ$ we obtain
$\nabla_T S=dS(T)+(\sigma^*ω)(T)\cdot S$
Now my question is if $ϕ$ is a function a $0$ form then $S$  the pullback of a $0$ form is a function but by definition of covariant derivative we should have
$\nabla_T S=dS(T)$ 
Only if if we identify $S$ as a section that we would have the extra piece.
Then how can we say that $\nabla_T$ is a covariant derivative?
 A: Okay, I still don't completely understand you, but I think I pinpointed the problem.
Using your notation, $S$ is not really a section of the associated bundle, but rather the component representation of such. So the notation $\nabla_T S$ is either incorrect, or just requires a lot of interpretation.
Let $F$ be a real, $k$ dimensional vector space, to see an example. Let us assume, that the latin indices $a$ range from $1$ to $k$.
We actually have some separate things here. We have a principal fiber bundle $(P,\pi,M,G)$ and an associated vector bundle $(E,\pi,M)$ with $E=P\times_\rho F$, where $\rho:G\rightarrow GL(F)$ the representation that defines the associated vector bundle.
The thing is, if $\psi:M\rightarrow E$ is a (smooth) section of $E$, then there exists a unique corresponding equivariant map $\phi:P\rightarrow F$, but for all intents and purposes, they are not the same.
You should instead think about them like this: If ${e_{a}}$ is a local frame of $E$, then we have locally $\psi=\psi^ae_a$, where the components $\psi^a$ depend on two things - they depend on the manifold points $x\in M$, but also on the local frame $e_a(x)$ (at $x$) chosen to represent it, so $$ \psi^a=\psi^a(x,\{e_a(x)\}). $$
But if $E$ is an associated vector bundle to $P$, then, essentially, $P$ is an associated principal bundle to $E$, so $P$ can be identified as the frame bundle of $E$ (because a local frame $e_a$ provides a local trivialization of both $E$ and $P$, so local sections of $P$ can be understood as frames of $E$). Well, not the total frame bundle, but some restriction of the frame bundle to a $G$-subbundle.
So the point is, that because the components $\psi^a$ depend on points of $P$, rather than $M$, it is an $F$-valued (essentially $\mathbb{R}^k$-valued) function on $P$. But it needs to be equivariant. Why? Because if $\Lambda^a_{\ b}(x)$ is a local frame transformation on $E$, then the components transform by $\Lambda^{-1}$. But a local frame transformation (acting on the model fiber $F$) is equivalent to the fibrewise transitive right action of $G$ on $P$, so $$ \psi^a(x,e(x)\Lambda(x))=(\Lambda^{-1})^a_{\ b}(x)\psi^b(x,e(x)).$$
Essentially, invariant fields are sections of $E$, component representations of it are realized basis independently as equivariant functions on $P$. Equivariance guarantees that the components transform "properly" under frame transformations (frame transformation = group action on $P$).

With this discussed, let us concatenate notation:


*

*$\psi$ is the invariant section of $E$.

*$\psi^a$ are components of it taken with respect to a local trivialization of either $E$ or $P$ (they agree).

*$\phi$ is the equivariant function on $P$ satisfying $$ \phi(x,e(x))=(\psi^1(x,e(x)),...,\psi^k(x,e(x))), $$ where $(x,e(x))$ is a point of $P$.
What actually is, is that $$ \nabla_T\psi=(d(\sigma^*\psi^a)(T)+(\sigma^*\omega)^a_{\ b}(T)(\sigma^*\psi^b))e_a, $$ where the frame $e_a$ is the frame that generates the local trivialization/section $\sigma$.
Since this got quite complicated, and the otherwise simple essence gets lost, what your "$\nabla_TS$" is, is an analogue of $\nabla_T X^\mu$ for tangent vectors, when, in fact, what interests you is not $\nabla_T X^\mu$, but $\nabla_TX=\nabla_T(X^\mu\partial_\mu)$.
You got the component representation of the covariant derivative. Which is fine, just the notation confuses you, because $\nabla_T$ should not act on the components, but the invariant section.
