Which derivative with respect to time is which in the Heisenberg picture of quantum mechanics?

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 $t$.

The Heisenberg picture is defined as

$$A_{\mathrm{H}}(t) = e^{iHt/\hbar} A_{\mathrm{S}}(t) e^{-iHt/\hbar}$$

differentiating both sides we obtain

$$i\hbar \frac{\mathrm{d}}{\mathrm{d} t} A_{\mathrm{H}}(t) = [ A_{\mathrm{H}}(t), H] + i\hbar \left( \frac{\mathrm{d}}{\mathrm{d} t} A_{\mathrm{S}}(t) \right)_{\mathrm{H}} \>\>\>\>\>\>\>\>\>\>\>\>\>\> (1)$$

Some textbooks rewrite the last term using the notation [*]

$$\frac{\partial}{\partial t} A_{\mathrm{H}}(t) \equiv \left( \frac{\mathrm{d}}{\mathrm{d} t} A_{\mathrm{S}}(t) \right)_{\mathrm{H}}$$

[*] I agree on that this notation is awkward for mathematicians (it is not a true partial derivative) and the more rigorous physics textbooks use (1) with the total time derivative.

It's easiest to derive this from the Schrödinger picture:

Let $B(t)$ be a time-dependent operator in the Schrödinger picture. The corresponding operator in the Heisenberg picture is $A(t) = e^{iHt/\hbar} B(t) e^{-iHt/\hbar}$. Differentiation with respect to $t$ gives

$$ \frac{d}{dt} A(t) = e^{iHt/\hbar} \left(\frac{i}{\hbar} H B(t) + \frac{\partial}{\partial t}B(t) - \frac{i}{\hbar} B(t) H) \right) e^{-iHt/\hbar} $$ $$ = e^{iHt/\hbar} \left(\frac{i}{\hbar} [H,B(t)] + \frac{\partial}{\partial t}B(t)\right) e^{-iHt/\hbar} = \frac{i}{\hbar} [H,A(t)] + \frac{\partial A}{\partial t} $$

In other words, the last partial derivative is to be understood in the sense that you take the operator $\frac{\partial B}{\partial t}$ and "evolve it in time" via the Schrödinger equation.

Useful non-example: the velocity operator $\vec v$. The velocity operator is the derivative of the position operator, but it's the total derivative as the system evolves. Hence,

$$ \vec v = \frac{i}{\hbar} [H,\vec r] .$$

In the Schrödinger picture, the position operator is, of course, time independent. Since $H$ is time independent as well, this is also the right velocity operator in the Schrödinger picture.