You have to go back to see where your equation $(1)$ came from.
For one dimension the probability of particles having a velocity between $\vec v_{\rm x}$ and $\vec v_{\rm x}+ d\vec v_{\rm x}$ is given by
$$f(\vec v_{\rm x})\;d\vec v_{\rm x} =\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,d\vec v_{\rm x}$$
The speed distribution is given by
$$f(v_{\rm x})\;dv_{\rm x} =2\,\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}=\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}$$
the factor $2$ being there because the speed (magnitude) of $\vec v_{\rm x}$ is the same as that of $-\vec v_{\rm x}$.
This is in agreement with your equation $(3)$.
Equation $(1)$ came from the idea that in three dimensions there is no preferred direction so
$f(\vec v_{\rm x},\vec v_{\rm xy},\vec v_{\rm z})\,d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z} = f(\vec v_{\rm x})d\vec v_{\rm x}\,f(\vec v_{\rm y})d\vec v_{\rm y}\,f(\vec v_{\rm z})d\vec v_{\rm z}$
You now have to count all the speeds which are the same ie the magnitude of the velocity $v$ is the same where $v^2 = v^2_{\rm x}+v^2_{\rm y}+v^2_{\rm z}$.
In this three dimensional case the volume of a shell of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z}= 4 \pi v^2 dv$ is being considered which results in your equation $(1)$.
$$f(v) \,dv = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
The equivalent distribution for two dimensions with an area of a ring of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}= 2 \pi v\, dv$ and $v^2 = v^2_{\rm x}+v^2_{\rm y}$ is
$$f(v) \,dv = \left(\frac{m}{2 \pi kT}\right)\, 2\pi v\, \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
Kinetic energy distribution $f_E(E)$ is a function that gives probability per unit energy interval.
Since kinetic energy $E$ is a function of speed, this distribution is related to speed distribution $f_v(v)$, but its value at any energy $E$ is not merely proportional to value of $f_v(v(E))$. The derivation of $f_E(E)$ from $f_v(v)$ relies on the substitution theorem from the integral calculus.
For positive speeds $v_1,v_2$ and corresponding energies $E_1=\frac{1}{2}mv_1^2, E_2=\frac{1}{2}mv_2^2$, the probability that particle has kinetic energy in the interval $(E_1,E_2)$ is the same as the probability that it has speed in the interval $(v_1,v_2)$. We transform the expression of the second probability in the following way:
$$
P = \int_{v_1}^{v_2}f_v(v)dv => (subst. theorem) => \int_{E_1}^{E_2}f_v(v(E))v'(E) dE
$$
where
$$
v(E) = (2E/m)^{1/2}
$$
and $v'(E)$ is derivative of this function at value $E$.
The expression of the first probability is
$$
P = \int_{E_1}^{E_2}f_E(E)dE.
$$
Since we can choose $v_1,v_2$ arbitrarily close to each other, it must be
$$
f_E(E) = f_v(v(E))\frac{dv(E)}{dE}.
$$
Substituting
$$
f_v(v) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2k_B T}}
$$
we obtain
$$
f_E(E) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} 4\pi 2E/m \cdot e^{-\frac{E}{k_B T}} \cdot (2/m)^{1/2}\frac{1}{2}E^{-1/2}
$$
and after simplification
$$
f_E(E) = \left(\frac{1}{\pi k_BT}\right)^{3/2} 2\pi \cdot E^{1/2} e^{-\frac{E}{k_B T}}.
$$
Best Answer
The Boltzmann distribution applies to each component of the energy.
For instance, in your monoatomic gas can be either in the fundamental electronic state or in an excited state an energy $E_{exc}$ above the fundamental one, in addition to a kinetic energy
$$E_k=m(v_x^2+v_y^2+v_z^2)/2$$
then the distribution function for an atom in the excited state will be proportional to
$$e^{-\frac {E_{exc}+E_k} {k_B T}}$$
while that in the fundamental will just be
$$e^{-\frac {E_k} {k_B T}}$$
If you ignore the excitation state, and consider only the component of the velocity along $x$, that is, you add all the atoms with a given $v_x$ whatever the degree of excitation or the values of $v_y$ and $v_z$ are, of course then the only relevant part is
$$e^{-\frac {mv_x^2} {2k_B T}}$$