Different formulas for $\rm SO(3)$ rotations For $\rm SO(3)$ rotations, the group elements are given by the standard Euler matrices $R_x(\theta_x)$, $R_y(\theta_y)$ and $R_z(\theta_z)$ for rotations in 3D space:
$$R_x(\theta_x)=\begin{bmatrix}
1 &0 &0 \\ 0& \cos\theta_x &\sin\theta_x \\0 &-\sin\theta_x & \cos\theta_x
\end{bmatrix} 
 $$
$$R_y(\theta_y)=\begin{bmatrix}
\cos\theta_y & 0 & -\sin\theta_y \\ 0& 1 &0 \\ \sin\theta_y& 0& \cos\theta_y
\end{bmatrix}
$$
$$R_z(\theta)=\begin{bmatrix}
\cos\theta_z &\sin\theta_z & 0\\ -\sin\theta_z& \cos\theta_z &0 \\0 & 0& 1
\end{bmatrix}
$$
The corresponding generators $J_x$, $J_y$ and $J_z$ are defined such that $$R_x(\theta_x)=e^{i\theta_xJ_x} 
 , R_y(\theta_y)=e^{i\theta_yJ_y},R_z(\theta_z)=e^{i\theta_zJ_z}.$$
I then read that the general rotation transformation is given by
$$R(\vec{\theta})=e^{i\vec{\theta}\cdot\vec{J}},$$
where $\vec{J}=(J_x, J_y, J_z)$.
I am confused about what the components for $\vec{\theta}$ will be.
Consider a rotation in space where we first rotate by $\theta_z$, then by $\theta_y$ and lastly $\theta_x$, the rotation matrix should be
$$R(\theta_x,\theta_y,\theta_z)=R_x(\theta_x)R_y(\theta_y)R_z(\theta_z)=e^{i\theta_xJ_x}e^{i\theta_yJ_y}e^{i\theta_zJ_z}.$$
Intuitively I would think that using the $R(\vec{\theta})=e^{i\vec{\theta}\cdot\vec{J}}$ formula means using $\vec{\theta}=(\theta_x, \theta_y, \theta_z)$:
$$R(\vec{\theta})=e^{i\theta_xJ_x+i\theta_yJ_y+i\theta_zJ_z}$$
However, as the generators $J_x, J_y, J_z$ don't commute, $$e^{i\theta_xJ_x}e^{i\theta_yJ_y}e^{i\theta_zJ_z} \neq e^{i\theta_xJ_x+i\theta_yJ_y+i\theta_zJ_z}.$$
So it is wrong to say that $\vec{\theta}=(\theta_x, \theta_y, \theta_z)$. What then should the components for $\vec{\theta}$ be?
 A: Your reasoning is correct.  The Euler angles are not the components of $\vec{\theta}$.  Here is what $\vec{\theta}$ is.
Let $\vec{\theta}=(\theta_1,\theta_2,\theta_3)=\theta \hat{n}$. Let's derive the 3x3 matrix (ie: the group elements $R(\vec{\theta})$) for rotating an object by $\theta$ radians about an arbitrary direction specified by the unit vector $\hat{n}$. This means put your right hand thumb along the unit vector $\hat{n}$ and rotate the object by pushing with your right hand fingers through the angle $\theta$.  For me, this is a much easier way to parameterize and visualize an arbitrary rotation than Euler angles.  Notice $\theta=\sqrt{\theta_1^2+\theta_2^2+\theta_3^2}$ .
As you say in your question, the group element is $R(\vec{\theta})=e^{i\vec{\theta}\cdot \vec{J}}$. This rotates any object which includes vectors with any number of components (for example, an arrow, a rock, a tensor, or particles with different spins).  We want to rotate a 3-vector so we put in the 3x3 matrix representation of each of the 3 generators $\vec{J}=(J_1,J_2,J_3)$.
$$
\begin{align}
\Theta & =i\vec{\theta}\cdot \vec{J} \\
       & = i\theta_1\begin{bmatrix}
        0 &         0 &         0 \\
        0 &         0 &         i \\ 
        0 &        -i &         0 \\
 \end{bmatrix}
+i\theta_2\begin{bmatrix}
        0 &         0 &        -i \\
        0 &         0 &         0 \\ 
        i &         0 &         0 \\
 \end{bmatrix}
+i\theta_3\begin{bmatrix}
        0 &         i &         0 \\
       -i &         0 &         0 \\ 
        0 &         0 &         0 \\
 \end{bmatrix} \\
     & = \begin{bmatrix}
        0 & -\theta_3 &  \theta_2 \\
 \theta_3 &         0 & -\theta_1 \\ 
-\theta_2 &  \theta_1 &        0  \\
 \end{bmatrix}
\end{align}
$$
Notice that $[J_1,J_2]=iJ_3$ which is correct for rotation generators (=angular momentum).
Finally we expand $e^{\Theta}$ in a power series and matrix multiply $\Theta$ 's together to calculate each term.  You will find $\Theta^3=-\theta^2\Theta$.
$$
\begin{align}
R(\Theta) & =e^{\Theta} \\ & =I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +… \\
  & =I+\Theta(1-\frac{\theta^2}{3!}+\frac{\theta^4}{5!}-...)+\Theta^2(\frac{1}{2!}-\frac{\theta^2}{4!}+\frac{\theta^4}{6!}-...) \\ \\
R(\Theta) & =I+ \frac{\Theta}{\theta}sin(\theta)+\frac{\Theta^2}{\theta^2}(1-cos(\theta))
\end{align}
$$
This $R(\Theta)$ is the matrix for rotating any 3-vector about an arbitrary unit vector $\hat{n}$ by angle $\theta$. As an example, suppose $\hat{n}=(0,0,1)$, which is a rotation about the z-axis by theta.  Then the final equation for $R$ yields the familiar rotation matrix
$$
R(\Theta) = \begin{bmatrix}
   cos(\theta) & -sin(\theta) & 0 \\
   sin(\theta) &  cos(\theta) & 0 \\ 
             0 &            0 & 1 \\
 \end{bmatrix} 
$$
Notice that my $sin(\theta)$ is the opposite sign as yours because I am doing an active transformation on the object, whereas your formula is for a passive transformation on the coordinate axis (ie: $\vec{\theta}_{passive}=-\vec{\theta}_{active}$) .
A: take the Taylor series for  this rotation matrix:
$$R_x(\theta_1)=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( 
\theta_{{1}} \right) &-\sin \left( \theta_{{1}} \right) 
\\ 0&\sin \left( \theta_{{1}} \right) &\cos \left( 
\theta_{{1}} \right) \end {array} \right] 
$$
you obtain
$$R_x(\theta_1)=\left[ \begin {array}{ccc} (1)&0&0\\ 0&(1-{\frac {1
}{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{\theta_{{1}}}^{4}+O \left( {
\theta_{{1}}}^{6} \right) )&(-\theta_{{1}}+{\frac {1}{6}}{\theta_{{1}}
}^{3}-{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{6}
 \right) )\\ 0&(\theta_{{1}}-{\frac {1}{6}}{\theta_{
{1}}}^{3}+{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{
6} \right) )&(1-{\frac {1}{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{
\theta_{{1}}}^{4}+O \left( {\theta_{{1}}}^{6} \right) )\end {array}
 \right]
\tag 1$$
the rotation matrix $R_x$ is also
$$R_x=\text{exp}(i\,\theta_1\tau_1)$$
take the Taylor series
$$R_x=I_3+x+\frac 12 x\,x+\frac 16 x\,x\,x+\ldots$$
with $x=i\,\theta_1\,\tau_1$
and:
$$\tau_1=-i\,\left[ \begin {array}{ccc} 0&0&0\\ 0&0&1
\\ 0&-1&0\end {array} \right]
$$
thus:
$$R_x(\theta_1)=\left[ \begin {array}{ccc} (1)&0&0\\ 0&(1-{\frac {1
}{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{\theta_{{1}}}^{4}+O \left( {
\theta_{{1}}}^{6} \right) )&(-\theta_{{1}}+{\frac {1}{6}}{\theta_{{1}}
}^{3}-{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{6}
 \right) )\\ 0&(\theta_{{1}}-{\frac {1}{6}}{\theta_{
{1}}}^{3}+{\frac {1}{120}}{\theta_{{1}}}^{5}+O \left( {\theta_{{1}}}^{
6} \right) )&(1-{\frac {1}{2}}{\theta_{{1}}}^{2}+{\frac {1}{24}}{
\theta_{{1}}}^{4}+O \left( {\theta_{{1}}}^{6} \right) )\end {array}
 \right]
\tag 2$$
analog for the rotation matrix $R_y~$ and $R_z$
the rigid body rotation matrix is now:
$$R_x(\theta_1)\,R_y(\theta_2)\,R_z(\theta_3)\mapsto
\text{e}^{(i\,\theta_1\,\tau_1)}\,\text{e}^{(i\,\theta_2\,\tau_2)}\,\text{e}^{(i\,\theta_3\,\tau_3)}$$
with:
$$\tau_2= -i\,\begin{bmatrix}
   0 & 0 & -1 \\
   0 & 0 & 0 \\
   1 & 0 & 0 \\
 \end{bmatrix}~,\tau_3=-i\,\begin{bmatrix}
   0 & 1 & 0 \\
   -1 & 0 & 0 \\
   0 & 0 & 0 \\
 \end{bmatrix}$$
