The generalization of Newton's 2nd law to general relativity is given by

$$ m \frac{d^2 x^\mu}{d\tau^2} + m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = f^\mu \qquad (\star)$$
where $\tau$ is the proper time along the particle's worldline, $f$ is the net 4-force acting on the particle, and $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols corresponding to your choice of coordinates. 

In inertial Cartesian coordinates, all of the $\Gamma$s are equal to zero which means that 
$$m \frac{d^2x^\mu}{d\tau^2} = f^\mu$$
If you pick different coordinates in which the $\Gamma$s are *not* zero, then obviously you're going to need to include that extra term.  This happens if you use polar coordinates, for example, but it **also** occurs when you use *accelerated* Cartesian coordinates.  In the latter case, what you can do is simply move the extra terms to the other side of the equation and call them  *pseudoforces*:
$$m \frac{d^2 x^\mu}{d\tau^2} = f^\mu - \underbrace{m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau}}_{\text{pseudoforce}}$$

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Example: The coordinates which correspond to a relativistic observer undergoing constant proper acceleration are the [Rindler coordinates](https://en.wikiversity.org/wiki/Theory_of_relativity/Rindler_coordinates).  In this coordinate system, assuming proper acceleration $a_0$ along the $x$-axis, the line element becomes
$$ds^2 = -\left(1+ \frac{a_0 x}{c^2}\right)^2 c^2dt^2 + dx^2+dy^2+dz^2$$
The nonzero Christoffel symbols are
$$\Gamma^0_{10} = \Gamma^0_{01} =  \frac{a_0/c^2}{1+a_0 x/c^2}\qquad \Gamma^1_{00} = \left(1+\frac{a_0 x}{c^2}\right)a_0/c^2$$
which means that the relativistic generalization of Newton's 2nd law for $x$ is

$$ m \left(\frac{d^2x}{d\tau^2} + a_0 \frac{1+\frac{a_0 x}{c^2}}{\left(1+ \frac{a_0 x}{c^2}\right)^2 - \frac{v^2}{c^2}}\right) = f_x$$

In the nonrelativistic limit, this becomes
$$m \left(\frac{d^2 x}{dt^2} + a_0\right) = f_x \iff m \frac{dx^2}{dt^2} = f_x \underbrace{ -\ m a_0}_{\text{pseudoforce}}$$

which is exactly what we'd need to do in Newtonian mechanics if we wanted to switch to an non-inertial frame accelerating with $\mathbf a = a_0 \hat x$.


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So in summary, you have the following options:

 1. Choose a coordinate system in which all the $\Gamma$s vanish - that is, a global inertial frame. Such frames generally do not exist in curved spacetime, so you can only do this in the context of SR.
 2. Recognize and accept that the left-hand side of $(\star)$ has additional terms due to the $\Gamma$s, which reflect the fact that your coordinate basis changes with position.  If you've ever constructed Newton's laws in polar coordinates, this is what you do.
 3. Move the term $m\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau}$ to the right-hand side of the equation and call it a pseudoforce.  If you've ever constructed Newton's laws in a non-inertial frame, this is what you probably did.

In general curved spacetime, you can't escape the $\Gamma$s so your options are limited to 2 and 3. When the Wiki article says that you don't *need* pseudoforces, it means that you can choose option 2, not that you can ignore them outright.

Finally, though in GR you cannot make the $\Gamma$s vanish *everywhere*, you can *always* make an instantaneous choice of coordinates to make them vanish at a *point*.  As a result, at any given instant of time you can choose coordinates such that $f^\mu_{\mathrm{pseudo}} \equiv -m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = 0$. Because $f^\mu_{\mathrm{pseudo}}$ can be set to zero by a coordinate transformation, we know that it is **not** a 4-vector because if a 4-vector vanishes in one coordinate system, it must vanish in all coordinate systems. This is the *mathematical* distinction between real forces, which are 4-vectors and therefore coordinate-independent geometrical objects, and *pseudoforces*, which can be viewed as artifacts of your choice of coordinates.