Derivation of four-force We have two equivalent expressions for the four-momentum,
$$\textbf{P} = \gamma m (c,\vec{u}) \tag{1}\label{1}$$
$$\textbf{P} = (E/c,\vec{p}) \tag{2}\label{2}$$
where $\gamma = 1/\sqrt{1-(\frac{u}{c})^2}$. The four-force and the three-force are given by,
$$\textbf{F} = \frac{d \textbf{P}}{d\tau} \tag{3}\label{3}$$
$$\vec{f} = \frac{d \vec{p}}{dt} \tag{4}\label{4}$$
so that the four-force can be written as,
$$\textbf{F} = \frac{d}{d\tau}(E/c,\vec{p}) = \gamma \Big(\frac{dE}{cdt},\frac{d\vec{p}}{dt}\Big) = \gamma \Big(\frac{\vec{f} \cdot \vec{u}}{c},\vec{f}\Big)\tag{5}\label{5}.$$
There are several ways to derive,
$$\frac{dE}{dt} = \vec{f} \cdot \vec{u} \tag{6}\label{6}$$
to which I won't discuss the other ways on how to do it. I'm going to derive it the direct way, using the relativistic energy $E = \gamma mc^2$. So, $\frac{dE}{dt} = mc^2 \frac{d\gamma}{dt}$ (in this post I'm assuming a constant mass).
$$\frac{d\gamma}{dt} = \frac{\gamma^3}{c^2} u \frac{du}{dt} = \frac{\gamma^3}{c^2} \vec{u} \cdot \frac{d\vec{u}}{dt} \tag{7}\label{7}$$
$$\frac{dE}{dt} = \gamma^3 m \vec{u} \cdot \frac{d\vec{u}}{dt} \tag{8}\label{8}$$
The question now is how can we establish $\gamma^3 m \frac{d\vec{u}}{dt} = \vec{f}$? We need to have a form $\gamma^3 m \frac{d\vec{u}}{dt} = \frac{d(\gamma m\vec{u})}{dt} = \vec{f}$. So we study this form,
$$\frac{d(\gamma m\vec{u})}{dt} = \gamma m \frac{d\vec{u}}{dt} + m\vec{u} \frac{d\gamma}{dt} = \gamma m \frac{d\vec{u}}{dt} + \frac{\gamma^3 m}{c^2} \vec{u} \Big(\vec{u} \cdot \frac{d\vec{u}}{dt}\Big). \tag{9}\label{9}$$
Using vector algebra identity $A \times (B \times C) = (A \cdot C)B - (A \cdot B)C$, where $A=B=\vec{u}$ and $C=\frac{d\vec{u}}{dt}$,
$$\frac{d(\gamma m\vec{u})}{dt} = \gamma m \frac{d\vec{u}}{dt} + \frac{\gamma^3 m}{c^2} \vec{u} \Big(\vec{u} \cdot \frac{d\vec{u}}{dt}\Big) = \gamma m \frac{d\vec{u}}{dt} + \frac{\gamma^3 m}{c^2} \Bigg[(\vec{u} \cdot \vec{u}) \frac{d\vec{u}}{dt} + \vec{u} \times \Big(\vec{u} \times \frac{d\vec{u}}{dt}\Big)\Bigg]. \tag{10}\label{10}$$
If I assume $\vec{u} \times \frac{d\vec{u}}{dt} = 0$, then
$$\frac{d(\gamma m\vec{u})}{dt} = \gamma m \frac{d\vec{u}}{dt} + \frac{\gamma^3 m}{c^2} (\vec{u} \cdot \vec{u}) \frac{d\vec{u}}{dt} = \gamma m \frac{d\vec{u}}{dt} + \frac{\gamma^3 m}{c^2} u^2 \frac{d\vec{u}}{dt} = \gamma^3 m \frac{d\vec{u}}{dt}, \tag{11}\label{11}$$
the calculation of the equality in $\eqref{11}$ is plain algebra. Thus we have confirmed the form we needed in the statement above $\eqref{9}$. However, this can only be true if the three-force and velocity are completely aligned, but $\eqref{6}$ is a general form which does not say anything about the orientation of $\vec{f}$ and $\vec{u}$. Am I missing something here or is this really the case?
 A: Your calculation is totally correct, but you have one small error in your argument. First of all, you correctly calculated that
$$\frac{\mathrm{d}E}{\mathrm{d}t}=mc^{2}\frac{\mathrm{d}\gamma}{\mathrm{d}t}=\gamma^{3}m\bigg (\vec{u}\cdot\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\bigg ).$$
In the end you want to show that
$$\frac{\mathrm{d}E}{\mathrm{d}t}=\gamma^{3}m\bigg (\vec{u}\cdot\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\bigg )\stackrel{!}{=}\vec{f}\cdot\vec{u}.$$
Now the error lies in the following step of yours: You said in order for the above equality to be true, we must have that
$$\gamma^{3}m\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\stackrel{!}{=}\vec{f}$$
and this is what you tried to proof. However, this is not true. In general, if you have three vectors $\vec{a},\vec{b},\vec{c}\in\mathbb{R}^{3}$, then $\vec{a}\cdot\vec{c}=\vec{b}\cdot\vec{c}$ does in general NOT imply that $\vec{a}=\vec{b}$. You can easily construct counterexamples. Of course, if you would have that $\vec{f}=\gamma^{3}m\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}$ then the above claim is true, but again, the converse is in general not true, since you are not allowed to "cancel" in inner products.
To complete your proof, let us start again with the claim
$$\frac{\mathrm{d}E}{\mathrm{d}t}=\gamma^{3}m\bigg (\vec{u}\cdot\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\bigg )\stackrel{!}{=}\vec{f}\cdot\vec{u}$$
In order to proof this, we just have to calculate the right-hand side, which is, following your correct formula (9), given by
$$\vec{f}\cdot\vec{u}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}\cdot\vec{u}=\gamma m\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\cdot\vec{u}=\bigg [\gamma m\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}+\frac{\gamma^{3}m}{c^{2}}\vec{u}\cdot\bigg (\vec{u}\cdot\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\bigg )\bigg ]\cdot\vec{u}=...$$
As a next step, we can use your vector identity for the second term, so basically your step (10), but now with the difference that we have the addtional $\vec{u}$ on the right-hand side:
$$...=\gamma m\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\cdot\vec{u}+\frac{\gamma^{3}m}{c^{2}}\bigg [(\vec{u}\cdot\vec{u})\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}+\vec{u}\times\bigg (\vec{u}\times\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\bigg )\bigg ]\cdot\vec{u}$$
Now you see that the latter term always vanishes, since $\vec{u}\times\text{something}$ is perpendicular to $\vec{u}$ and hence
$$\underbrace{\bigg[\vec{u}\times\underbrace{\bigg (\vec{u}\times\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\bigg )}_{\text{something}}\bigg ]}_{\perp\vec{u}}\cdot\vec{u}=0$$
But again, for this to work you need the  "$\cdot\vec{u}$" from the above formula $(\vec{f}\cdot\vec{u})$, since, as you said correctly, $\vec{u}$ is in general not parallel to $\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}$ and hence, $[\vec{u}\times(\vec{u}\times\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}) ]$ does in general not vanish.
Now you can proceed as in your equation (11), which shows that
$$\vec{f}\cdot\vec{u}=\gamma^{3}m\bigg (\frac{\mathrm{d}\vec{u}}{\mathrm{d}t}\cdot\vec{u}\bigg )$$
and hence proves your claim.
