I could find the following satifactory answer to my question with the ideas provided in the comments above.

Consider the natural projection onto the projective plane, of a non-null vector into the line it spans

${\mathbb R}^3 - 0 \to {\mathbb P}^2 \qquad v \mapsto [v]$

and identify the tangent space $T{\mathbb P}^2_{[v]}$ with the corresponding tangent space of the sphere $S^2_v = v^\perp$.

For part (i), consider the chord map between $f$ and $g$, given by

$c = c(f,g): S^1 \times S^1 \to {\mathbb P}^2 \qquad
c(x,y) = [f(x) - g(y)]$

defined for $(x,y)$ such that $f(x) \neq g(y)$. It can be shown that intersecting the regular values of $c(f,g), c(f,f), c(g,g)$ and removing the tangent directions $[f'], [g']$ we get a full set of directions $[v] \in R$ such that under orthogonal projection to $v$, the projected curves are regular, intersect transversely and self-intersect transversely.

For part (ii), consider the secant map between $f$ and $g$, given by

$s = s(f,g): S^1 \times S^1 \times S^1 \to {\mathbb P}^2 \times {\mathbb P}^2
\qquad
s(x,y,z) = (\,[f(x) - g(y)],\, [g(y) - f(z)]\,)$

defined for $(x,y,z)$ such that $f(x) \neq g(y) \neq f(z)$. To avoid triple points we must avoid directions $v$ such that $s(x,y,z) = ([v],[v]) = \Delta([v])$, where $\Delta: {\mathbb P}^2 \to {\mathbb P}^2 \times {\mathbb P}^2$ is the diagonal embedding.
Note that we want to avoid directions whose image under $\Delta$ are in
$$
{\rm Im}\,s \cap \Delta({\mathbb P}^2) = s( s^{-1}( \Delta({\mathbb P}^2) ) )
$$
and that the image $\Delta({\mathbb P}^2)$ is an embedded submanifold of ${\mathbb P}^2 \times {\mathbb P}^2$. If the secant map was transversal to it, then $s^{-1}( \Delta({\mathbb P}^2) )$ would be a curve, so that the above intersection would be a differentiable image of a curve into a surface, hence a null set of ${\mathbb P}^2$.

Unfortunately it can only be shown that the secant map is transversal to $\Delta(R)$. Nevertheless, the above argument is local and still follows. Indeed, since the secant is transversal to the point $\Delta(w)$ for $w \in R$ (meaning that $s(x,y,z) = \Delta(w)$ implies that the image of $s'(x,y,z)$ complements the image of $\Delta'(w)$ in $T({\mathbb P}^2\times{\mathbb P}^2)_{\Delta(w)})$, by the same proof involving the local form of immersions there exists a neighbourhood $U$ of $\Delta(w)$ in $\Delta( {\mathbb P}^2 )$ such that
$$
{\rm Im}\,s \cap U = s( s^{-1}( U ) )
$$
is a differentiable image of a curve into a surface, hence a null set of $\Delta({\mathbb P}^2)$. These open sets give an open cover of $\Delta(R)$. Since ${\mathbb P}^2$ is a manifold and thus its subsets are Lindelöf, we get an enumerable cover $U_i$ of $\Delta(R)$. Removing from $\Delta(R)$ the null sets ${\rm Im}\,s \cap U_i$, we get another full set $R'$ contained in $R$ such that
$$
\Delta(R') \cap {\rm Im}\, s = \emptyset
$$

Note that for links we must consider the construction above with the secant map $s(f,g)$ and also $s(f,f)$ and $s(g,g)$. This is since each $f$ and $g$ can have self-intersections in ${\mathbb R}^3$ which we do not want to get projected to new self-intersections of $f$ and $g$ in the plane.