In wiki and open stax it is said that Newton’s Third Law includes the fact that Newton's Third Law involves the concept of direction in its statement but in this answer it is said that the direction aspect is not given in the statement of third law and requires extra experimental ideas.
The particular reason I'm concerned about this is that we could derive the conservation of angular momentum if the third law does have the direction aspect in its statement, so does it or does it not?
$\mathbf{F_1} = - \mathbf{F_2}$ according to third law but the forces are not along the connecting line.
Newton was vague on this, in the Principia. He made a general statement without reference to the direction the force acts in, but then cited an example where the force action/reaction is along the line between the two bodies, making it seem like that aspect of the situation was also meant to be taken as the general case.
But, as to your other comment: no, that's not enough to ensure the conservation of angular momentum. Not even if the forces line up with the bodies. That's because there is a gap in Newton's Third Law. There's no corresponding action-reaction law for helical torque!
If two bodies have torque-at-a-distance along the axis separating the two, then where's the law that says the torque on one should be equal and opposite to the torque on the other? That's missing.
Bear in mind that there are no net forces involved in that interaction. This is particularly the case if the bodies, themselves, have no constituency, where the angular momentum being torqued is intrinsic, i.e. not arising from the orbital motions of any constituency. Nobody ever said that Newton's "corpuscles", for instance, couldn't have intrinsic angular momentum. And, there was no place where Newton postulated any such restriction, either.
So, it's a gap.
So, the best you can say is that the Third Law allows you to argue for the conservation of the orbital part of angular momentum: that part of angular momentum that arises from the center of mass motion of the body; but nothing has any bearing on the internal part of angular momentum, the part that is independent of the body's position.
In general, a body has eleven quantities that are conserved when in isolation: its mass $m > 0$, its energy $H$ and the three components of each of the vectors: angular momentum $$, linear momentum $$ and the "moving mass moment" $$. In the absence of internal energy, denoted here $U$, and internal angular momentum, denoted $$, one has the following decomposition: $ = m - t$ and $ = m$, which can also be used, in reverse, to define the velocity and center of mass position of a body by the formulae: $$ = \frac{}{m}, \hspace 1em = \frac{}{m} + t;$$ $ = ≡ ×$, which is referred to as the "orbital" angular momentum; and $H = T ≡ ½m||^2$, which is referred to as the kinetic energy of the body.
As is, the formulae cannot be consistently maintained, since they are not additive. One generally assumes that for any decomposition of a body into two parts, with respective values $\left(_i, _i, _i, H_i, m_i\right)$, for $i = 0, 1$, that there is additivity, except for a possible interaction energy $V$: $$\left(_i, _i, _i, H_i, m_i\right) = \left(_0 + _1, _0 + _1, _0 + _1, H_0 + H_1 + V, m_0 + m_1\right).$$
This is not consistent with the formulae $ = $ and $H = T$. Instead, the correct formulae, that respect upward-scalability and additivity are: $$ = + = × + = m× + , \hspace 1em H = T + U = \frac{||^2}{2m} + U = ½m||^2 + U,$$ where $$ and $U$ are, respectively, the internal angular momentum and internal energy of the body. Then - and only then - can one consistently uphold additivity, with the respective decomposition being given by: $$ = \frac{m_0_0 + m_1_1}{m_0 + m_1}, \hspace 1em = \frac{m_0_0 + m_1_1}{m_0 + m_1}, \hspace 1em m = m_0 + m_1, \\ = _0 + _1 + μ×, \hspace 1em U = U_0 + U_1 + ½μ||^2 + V $$ with a portion of the internal values arising from the relative motions of the two parts of the body, describable in terms of a fictitious body with the attributes $$ = _0 - _1, \hspace 1em = _0 - _1, \hspace 1em μ = \frac{m_0m_1}{m_0 + m_1}.$$
Moreover, provided that our part-whole decomposition respects the mass content of each component body (i.e. $dm_0/dt = 0$ and $dm_1 = 0$) then we may consistently assume that $$ = \frac{d}{dt},$$ since it scales up from parts to whole, as does $dm/dt = 0$. That is: $$\left\{\begin{matrix}_0 = \frac{d_0}{dt}\\_1 = \frac{d_1}{dt}\end{matrix}\right\} ⇒ = \frac{d}{dt}.$$ The same applies to the fictious body, in that case: $$\frac{d}{dt} = , \hspace 1em \frac{dμ}{dt} = 0.$$
A portion of $$ - but not necessarily all of it - may arise from the internal motions of a body's constituencies. Nothing says that $$ has to all be reducible the orbital angular momenta associated with the motions of constituencies. A similar observation applies to $U$.
Now, consider the case where the two bodies in question interact with one another. Let $$ be the force of body 0 on body 1, and $-$ the equal and opposite reaction force of body 1 on body 0. Assume also that the forces are aligned between their respective separations; i.e. that $$ is collinear with $$. Assume also that the entire body is free from external forces. Then, we may only conclude that: $$\frac{d}{dt}μ× = μ\frac{d}{dt}× + μ×\frac{d}{dt}.$$ For the first vector we have: $$μ\frac{d}{dt}× = μ× = .$$ For the second vector, we have $$\frac{d}{dt} = \frac{d_1}{dt} - \frac{d_0}{dt} = \frac{}{m_1} - \frac{-}{m_0} = \frac{m_0 + m_1}{m_0m_1} = \frac{}{μ}.$$ Thus, $$μ×\frac{d}{dt} = × = ,$$ with the last equality happening by our collinearity assumption.
We can also say that $d/dt = $, applying a similar argument, since $d/dt = $ and $d/dt = $. However, what can not say is that $$ is conserved! Instead, what we get is: $$\frac{d}{dt} = \frac{d}{dt} + \frac{d_0}{dt} + \frac{d_1}{dt} + \frac{d}{dt}(μ×) = \frac{d_0}{dt} + \frac{d_1}{dt}.$$
Unless we also postulate that the respective torques: $$_0 = \frac{d_0}{dt}, \hspace 1em _1 = \frac{d_1}{dt},$$ are equal and opposite, i.e.: $$_0 = -_1,$$ particularly along the axis collinear with $$, then we're blocked from concluding that $d/dt = $.
Not only does $$ need not all arise from the motion of a body's constituents, but it may also be the case that the body has no constituents (that it is elementary, or a "corpuscle" as Newton called it), but yet with $ ≠ $. There is no condition that excluded that possibility - especially not by virtue of any "definition" of the form $ = $, since such definitions are blocked on consistency grounds, alone. Since $ = + $ must be maintained as the correct formula - at least for composite bodies - then there is no reason to assume that this does not also apply to elementary, non-composite, bodies, as well and that they, too, have internal angular momenta $$ of an entirely intrinsic nature.
Regardless of whether that's the case or not, the fact remains: the law for equal and opposite torques, $_0 = -_1$, is missing. And that gap is made all the more stark (not less!) if (a) the torques are on an axis aligned with $$ and (b) the Third Law assumes that $$ is also aligned with $$, since the usual definition, arising from that for orbital angular momentum, would have the vectors $_0$ and $_1$ each being perpendicular to $$, if they're all non-zero.
