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As in the title, I was studying the proof (from Vector Analysis - Louis Brand) of this identity, however I do not completely understand all of the steps.

$$a times (b times c) = b(a cdot c) - c(a cdot b)$$

I also visited an excellent related post - do the BAC-CAB identity for vector triple product have some interpretation?, but I am still stuck. I reproduce the proof in the book here. I don't really understand, how did the author deduce $alpha=-lambda(textbf{v}cdottextbf{w}),beta=lambda(textbf{u}cdottextbf{w})$ and the basis steps. Any inputs, suggestions or tips to understand the proof would be incredibly helpful!

Proof.

The vector $(textbf{u} times textbf{v}) times textbf{w}$ is perpendicular to both $textbf{u}timestextbf{v}$ and therefore coplanar with $textbf{u}$ and $textbf{v}$.

$$(textbf{u}timestextbf{v})timestextbf{w}=alphatextbf{u}+betatextbf{v}$$

But, since $(textbf{u}timestextbf{v})timestextbf{w}$ is also perpendicular to $textbf{w}$,

$$(alphatextbf{u}+betatextbf{v})cdottextbf{w}=0$$

All numbers $alpha,beta$ that satisfy this equation must be of the form $alpha=lambda(textbf{v}cdottextbf{w}),beta=lambda(textbf{u}cdottextbf{w})$, where $lambda$ is arbitrary.

Thus, we have

$$textbf{u}times(textbf{v}timestextbf{w})=lambda{(textbf{u}cdottextbf{w})v-(textbf{v}cdottextbf{w})textbf{u}}$$

In order to determine $lambda$, we use a special basis in which $hat{i}$ is collinear with $textbf{u}$, $hat{j}$ is co-planar with $textbf{u, v}$; then

$$textbf{u}=u_{1}i,textbf{v}=v_{1}i+v_{2}j,textbf{w}=w_{1}i+w_{2}j+w_{3}k$$

On substituting these values, we obtain, after a simple calculation, $lambda=1$.

We therefore have important expansion formulas,

$$(textbf{u}timestextbf{v})timestextbf{w}=(textbf{u}cdottextbf{w})textbf{v}-(textbf{v}cdottextbf{w})textbf{u}$$

$$textbf{w}times(textbf{u}timestextbf{v})=(textbf{w}cdottextbf{v})textbf{u}-(textbf{w}cdottextbf{u})textbf{v}$$

If we let $alpha=lambda(vcdot w)$, then
$$(alpha u+beta v)cdot w=lambda(vcdot w)(ucdot w)+beta(vcdot w)=0\implies beta=-lambda(ucdot w)$$ as required.
$$(utimes v)times w=lambda((vcdot w) u -(ucdot w) v)tag 1$$
(by the way, I have accidentally ended up with a different $lambda$ to the one given in your derivation, since in that, they seem to have suddenly decided to find an expression for $utimes(vtimes w)$ instead of $(utimes v)times w)$ as we have been working with the whole time. I think my one fits in better.)

For the basis steps, we choose a basis of our choice, $$u=u_{1}i,v=v_{1}i+v_{2}j,w=w_{1}i+w_{2}j+w_{3}k$$
Compute the LHS of $(1)$, $$(u_1v_1(itimes i)+u_1v_2(itimes j))times(w_1i +w_2j+w_3k)=u_1v_2ktimes(w_1i+w_2j)\=u_1v_2w_1j-u_1v_2w_2i$$

Then compute the RHS of $(1)$, $$(v_1w_1+v_2w_2)u_1i-u_1w_1(v_1i+v_2j)=(v_2w_2u_1)i-(u_1w_1v_2)j$$

Thus $lambda=-1$, and hence $$(utimes v)times w=(ucdot w) v-(vcdot w) u tag 2$$
Equivalently $$w times (u times v) = u(v cdot w) - v(u cdot w)tag3$$and hence $$a times (b times c) = b(a cdot c) - c(a cdot b)tag4$$