There is no mistake on the Wikipedia page and all the equations and statements are consistent with each other. In
$$A_{\rm Heis.}(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$$
the letter $A$ in the middle of the product represents the Schrödinger picture operator $A = A_{\rm Schr.}$ that is not evolving with time because in the Schrödinger picture, the dynamical evolution is guaranteed by the evolution of the state vector $|\psi\rangle$.
However, this doesn't mean that the time derivative $dA_{\rm Schr.}/dt=0$. Instead, we have
$$ \frac{dA_{\rm Schr.}}{dt} = \frac{\partial A_{\rm Schr.}}{\partial t} $$
Here, $A_{\rm Schr.}$ is meant to be a function of $x_i, p_j$, and $t$. In most cases, there is no dependence of the Schrödinger picture operators on $t$ - which we call an "explicit dependence" - but it is possible to consider a more general case in which this explicit dependence does exist (some terms in the energy, e.g. the electrostatic energy in an external field, may be naturally time-dependent).
In Schrödinger's picture, $dx_{i,\rm Schr.}/dt=0$ and $dp_{j,\rm Schr.}/dt=0$ which is why the total derivative of $A_{\rm Schr.}$ with respect to time is given just by the partial derivative with respect to time. Imagine, for example,
$$ A_{\rm Schr.}(t) = c_1 x^2 + c_2 p^2 + c_3 (t) (xp+px) $$
We would have
$$ \frac{dA_{\rm Schr.}(t)}{dt} = \frac{\partial c_3(t)}{\partial t} (xp+px).$$
These Schrödinger's picture operators are called "untransformed" on that Wikipedia page. The transformed ones are the Heisenberg picture operators given by
$$A_{\rm Heis.}(t) = e^{iHt/\hbar} A_{\rm Schr.}(t) e^{-iHt/\hbar}$$
Their time derivative, $dA_{\rm Heis.}(t)/dt$, is more complicated. An easy differentiation gives exactly the formula involving $[H,A_{\rm Heis.}]$ that you quoted as well.
$$\frac{d}{dt} A_{\rm Heis.}(t) = \frac{i}{\hbar} [H, A_{\rm Heis.}(t)] + \frac{\partial A_{\rm Heis.}(t)}{\partial t}.$$
The two terms in the commutator arise from the $t$-derivatives of the two exponentials in the formula for the Heisenberg $A_{\rm Heis.}(t)$ while the partial derivative arises from $dA_{\rm Schr.}/dt$ we have always had. (These simple equations remain this simple even for a time-dependent $A_{\rm Schr.}$; however, we have to assume that the total $H$ is time-independent, otherwise all the equations would get more complicated.) The two exponentials on both sides never disappear by any kind of derivative, so obviously, all the appearances of $A$ in the differential equation above are $A_{\rm Heis.}$. The displayed equation above is the (only) dynamical equation for the Heisenberg picture so it is self-contained and doesn't include any objects from other pictures.
In the Heisenberg picture, it is no longer the case that $dx_{\rm Heis.}(t)/dt=0$ (not!) and the similar identity fails for $p_{\rm Heis.}(t)$ as well. $A_{\rm Heis.}(t)$ is a general function of all the basic operators $x_{i,\rm Heis.}(t)$ and $p_{j,\rm Heis.}(t)$, as well as time $t$.
Neither left-hand side nor right-hand side are properly operators that act on states and give states. They are a form of operators that act on states and give complex numbers. We call such a thing a bra (and a state is called a ket, so when you pair them you get a bra(c)ket).
The left-hand side acts like $$((H\psi)^*)(\varphi) = \int \psi^* H \varphi \, dx.$$
If this confuses you, think about the finite-dimensional case. Then $\psi$ is a column vector and $H$ is a square matrix. While there is a complex conjugation operation for vectors, what you really want is the Hermitian conjugate, which is transpose and complex conjugation, denoted with a ${}^\dagger$. Then $(H\psi)^\dagger$ is the transpose of a column vector, so it's a row vector, and $(H\psi)^\dagger = \psi^\dagger H$ and this can act on a column vector with matrix multiplication.
For wavefunctions, $(\psi^\dagger)(x) = \psi(x)^*$ where the star is complex conjugation, and matrix multiplication is of a row vector and a column vector is integration.
Best Answer
One comment pointed out that the derivative $\frac{\partial}{\partial t} \frac{d}{dx} = 0$. In a certain sense however $ \frac{d}{dt} \frac{d}{dx} \neq 0$. I will try to explain this further.
Let's think first about kinematics. Then we have a Hilbert space $\mathscr{H}$ and some operator $A$ on it. For example the Hilbert space could be the space of square-integrable functions and $A$ the derivative, $\mathscr{H} = L^2(\mathbb{R})$ and $A = \frac{d}{dx}$.
If we want to study time evolution, we have to prescribe an one-parameter group of unitaries, $U_t$. We may now ask for the orbits of operators under this group of unitaries, that is, we might be interested in
$$ A(t) := U_t^{-1} A U_t \ . $$
We might additionally study observables which we let parametrically depend on $t$; for example we may multiply an operator with a function $f(t)$. I will ignore this in the following, but it is not hard to take into account. Take a time evolution operator
$$U_t = \exp(- i H t) \ .$$
Then we may compute the time derivative of $A(t)$:
$$ \frac{d A(t)}{dt} = \frac{d }{dt} \left(e^{ i t H} A e^{-i t H}\right) = (i H) e^{ i t H} A e^{-i t H} + e^{ i t H} A e^{-i t H} ( - i H) = i [H,A(t)] \ . $$
Let's consider the case of $A = \frac{d}{dx}$ in more detail, and let's choose
$$H = - \frac{1}{2} \frac{d^2}{d x^2} + \frac{1}{2} x^2 \ ,$$
which is of course the harmonic oscillator. Now we want to see what
$$A(t) = \left(\frac{d}{dx}\right)(t)$$
is. Note that this is no longer the derivative! The connection it has to the derivative is that
$$ \left(\frac{d}{dx}\right)(0) = \frac{d}{dx} \ .$$
The equation of motion for $A$ is
$$ \frac{d A(t)}{dt} = i [ H(t), A(t)] = \frac{i}{2} \left[ x(t)^2, \left(\frac{d}{dx}\right)(t) \right] \ . $$
Here i used that $H(t) = H$. In this formula we may now use that for two operators $B(t),C(t)$ that are obtained from operators $B,C$ by conjugating with an unitary $U_t$, it holds that
$$ [B(t),C(t)] = ([B,C])(t) \ .$$
Hence
$$ \left[ x(t)^2, \left(\frac{d}{dx}\right)(t) \right] = \left(\left[ x^2, \frac{d}{dx}\right] \right)(t) = 2 x(t) \ . $$
Playing the same game with $x(t)$, we get the set of coupled equations
$$ \frac{d \left(\frac{d}{dx}\right)\!(t)}{d t} = i x(t) \ , \\ \frac{d x(t)}{dt} = i \left(\frac{d}{dx}\right)\!(t) \ .$$
which may be easily solved to give
$$ \left(\frac{d}{dx}\right)\!(t) = \cos(t) \frac{d}{dx} + i \sin(t) x \, \\ x(t) = \cos(t) x + i \sin(t) \frac{d}{dx} \ . $$