I still don't understand how a gas with an equation of state f(T,p,V)=0 can change if 2 state functions are fixed? Just like classical potential, it's stated that equilibrium is obtained when the corresponding thermodynamics potential reaches the minimum.
Explicitly, according to Wikipedia,
in particular:

When the entropy (S ) and "external parameters" (e.g. volume) of a closed system are held constant, the internal energy (U ) decreases and reaches a minimum value at equilibrium.
When the temperature (T ) and external parameters of a closed system are held constant, the Helmholtz free energy (F ) decreases and reaches a minimum value at equilibrium.
When the pressure (p) and external parameters of a closed system are held constant, the enthalpy (H ) decreases and reaches a minimum value at equilibrium.
When the temperature (T ), pressure (p) and external parameters of a closed system are held constant, the Gibbs free energy (G ) decreases and reaches a minimum value at equilibrium.

There is something confuse me here. Take the first claim for example. If we have the formula: $dU=TdS-pdV$ then when $S$ and $V$ are held constant, is it not clear that $dU=0$ for whatever changes occurring? Furthermore, if there are $two$ fixed parameter, is it not true that the state of system is fixed? (assuming there is a state formula $f(T,p,V)=0$).
It seems like I'm really misunderstanding some concepts of the thermodynamics potential. Can someone give me a clear explanation for my confusion?
 A: The fundamental thermodynamic relation, $\mathrm{d}U=T\mathrm{d}S−p\mathrm{d}V$, describes the system only at equilibrium. If the system is not at equilibrium initially and $S$ and $V$ are held constant, it will evolve toward the state where $U$ is minimum.
A: Note that the cited Wikipedia article refers to external parameters  What's unclear from the context is that there can be internal parameters which are free to vary.  For example, the location of a partition, particle density on either side of a semi-permeable membrane, temperature on either side of a conducting membrane ... that's just a few of endless possibilities.
The question is: With all the external parameter held fixed, what are the values of the internal parameters at equilibrium?   It is those values that are given by minimizing the appropriate potential.
A: It is a good question and your confusion is genuine, especially in view of the fact that you are trying to understand the meaning of the statements given in Wikipedia, that need a lot of elaboration.
But simple tings first.
If you try to apply the concepts/equations of the first and second law of thermodynamics to closed system containing gas governed by a state formula f(T,p,V)=0, there is no easy way for you to understand the concepts of Helmhotz and Gibbs free energy changes - they are connected with the more subtle concepts of availability and so on.
The answer to your question:
'Furthermore, if there are two fixed parameter, is it not true that the state of system is fixed?' is
'yes'; the state of the system is fixed without doubt, if two of the three parameters are fixed.
Coming to the other aspects connected with Wikipedia, the simplest way to understand them is to construct the verbal staements of the concerned equations. We take one by one, below.
1a. Internal energy (U ) is the capacity to do work plus the capacity to release heat. This is the verbal statement corresponding to the mathematical statement:
Change in internal energy = dU = (Q - W) 
We use the convension W > 0 for work done by the system and Q>0 is the heat supplied to the system.
1b. When the entropy (S ) and "external parameters" (e.g. volume) of a closed system are held constant, the internal energy (U ) decreases and reaches a minimum value at equilibrium.
This is not a correct statement. Internal energy change does not give a criterion either for equilibrium or for the direction of a process. This is the very reason that F and G have been formulated in thermodynamics. They give the criteria for equilibrium of a system and spontaniety of a process
dU = (Q - W)
= TdS - W (W = pdV + other forms of work).
dU = 0 at const S and constant external parameters (V for example)*
Therefore, U is constant when the system suffers no heat interaction and work interaction with the surroundings.
*Using V as an external parameter is not a good practice, we use it here since Wiki used it like that and we are discussing wiki notes. 
For gases, the only work is the pdV work. But, for systems such as chemical reactions, other forms of work such as electrical work come into picture.
Therefore, for arbitrary (general) system the 1st law is:dU = Q - W
and, for gases the first law is dU = Q - pdV 
For reversible processes we get:dU = Qrev - Wmax (or -Wmin, if work is done on the system).
The second law statement is dQ ≥ TdS (equlity applies for rev processes)
The combined statement of the first and second laws for rev processes is:
dU = (TdS - Wmax).
This eqn combined with the definitions of H,F and G, gives all the statements of Wikipedia.
2a. Enthalpy is the capacity to do non-mechanical work plus the capacity to release heat. This is the verbal form of the equation:
H = U + pV + XY (X is an intensive and Y is an extensive property of the syatem.
2b. ΔH is the sum of non-mechanical work done on the system and the heat given to it.
∆H = Q(p) + XdY (at const p and const X), obtained as follows.
H = U + pV + XY 
∆H = ∆U + ∆(PV) + ∆(XY) = Q - pdV + pdV + V dp + YdX + non mechanical (pdV) work.
∆H = Q(p) + XdY (at const P and const X) 
The above equation is the mathematical form of the verbal statement:
Increase in Enthalpy (∆H) at const p, and const external conditions,  is the heat supplied to the system plus the work other than pdV work supplied to the system.
F and G
U and H are energies that fail to give criterion for equilibrium or the direction of a process, which is the prime concern of thermodynamics. Second law gives such a criterion: dS ≥ Q/T, but the presence of Q which is not a property of the system makes it not a very useful criterion. We look for a criterion that gives the direction of a process from the change in properties of the sysyem alone. F and G are such criterion. 
F = E - TS; G = H - TS
3a. Criteria for spontaneity and equilibrium in terms of F.
When the temperature (T ) and external parameters of a closed system are held constant, the Helmholtz free energy (F ) decreases and reaches a minimum value at equilibrium. This statement is the verbal form of the mathematical statement:
At const T, ∆F = ∆U - T∆S = (Q - W) - T∆S. Under const External conditions of W=0. ∆F = Q - T∆S, and dS ≥ Q/T
When temperature and external parameters are held constant(W = 0),  we get,
-dF ≥ TdS - TdS = 0, Therefore, F is an extreamum, minimum for equilibrium.
The inequality gives the criterion for a spontaneous process and the equality gives the condition for equilibrium.
3b. Change in F corresponds to maximum work possible.
Wiki says 'Helmholtz free energy is the capacity to do mechanical work (useful work)'. Note, mechanical work applies to systems of gas only, where the work can only be just pdV work. 
The above statement is the verbal form corresponding to the eqn 
-∆F = Wmax, obtained as follows
At const T, ∆F= ∆U -T∆S = Qrev - Wmax - T∆S = T∆S - Wmax - T∆S = - Wmax. 
4a. Criteria for spontaneity and equilibrium in terms of G.
When the temperature (T ), pressure (p) and external parameters of a closed system are held constant, the Gibbs free energy (G ) decreases and reaches a minimum value at equilibrium. This statement is the verbal form of the mathematical statement:
∆G = ∆U + pdV - T ∆S = (Q - W) + pdV - T∆S
At const T, p and under const External parameters of W = 0,
∆G = Q -TdS, and dS ≥ Q/T. 
The inequality gives the criterion for a spontaneous process and the equality gives the condition for equilibrium.
Therefore, the criterion for spontaneous change is ∆G < 0,
and, the criterion for equilibrium is ∆G = 0, G is an extreamum, minimum for equilibrium.
4b ΔG is the non-mechanical work done on the system
The above statement is the verbal form corresponding to the eqn 
-∆G = Wmax - pdV , obtained as follows:
At const T, ∆G = ∆H - T∆S = ∆U + pdV - T∆S = ∆F + pdV = - (Wmax -pdV) = -Wnet. 
A: Let $T_0$ and $p_0$ denote the temperature and pressure of the environment and let us put some system in this environment. Then, using the Gibbs equation, $\mathrm{d} U= T \mathrm{d} S - p \mathrm{d}V$, it is easy to verify that:
$$
T_0 \mathrm{d} S_0 = - \left( \mathrm{d} U + p_0 \mathrm{d} V  \right)
$$
Furthermore, let $\mathrm{d} S$ denote the change in entropy of the system, such that the total change of entropy is:
$$
\mathrm{d} S_{\text{total}} = \mathrm{d} S_0  + \mathrm{d} S \geq 0
$$
Hence:
$$
T_0 \mathrm{d} S_{\text{total}} = - (\mathrm{d} U + p_0 \mathrm{d} V -  T_0 \mathrm{d} S) \geq 0 
$$
or equivalently:
$$
\mathrm{d} U + p_0 \mathrm{d} V - T_0 \mathrm{d} S \leq 0 \tag{1}
$$
This inequality play an important role when systems are moving to equilibrium.
In order to see this let us consider a system with constant entropy and volume, i.e. $\mathrm{d}S= \mathrm{d} V =0$. Then equation $(1)$ becomes:
$$
\mathrm{d} U  \leq 0 
$$
which implies that $U$ decreases and reaches a minimum value at equilibrium. In exactly the same way, you should be able to understand the other examples you have quoted.

Furthermore, if there are two fixed parameter, is it not true that the state of system is fixed? (assuming there is a state formula $f(T,p,V)=0)$.

The equation of state $f(T,p,V)=0)$ tells us that if we know, say $T$ and $p$, then we also know $V$. In other words, one of the three variables $T$, $p$ or $V$ (whichever one you choose) can be expressed in terms of the other two. Thus, the state of the system is completely determined by two out of the three quantities $T$, $p$ and $V$. I really don't understand what this has to do with the example you are referring to and why this is confusing you. If we don't change the volume and entropy, then we still have not exhausted the state variables.
