This is a very loose answer since it's based on something that, as far as I know, has not been fully formalized. The key idea is that of what I might call a "dependent arrow" between two objects.
The closest analogy is to that of a dependent path in HoTT, where if we have $a, a' : A$, $p : a = a'$, $b : B(a)$ and $b' : B(a')$, then even though $b$ and $b'$ have different types, we can still talk about paths between them using $p$. Specifically, the type of dependent paths is $p \# b = b'$, where $\#$ denotes transportation. As you probably, know, if $f : \Pi_{a : A} B(a)$, then $f$ maps a path $p : a = a'$ to a dependent path $apD_f(p) : p \# f(a) = f(a')$.
In a hypothetical directed (homotopy) type theory, we might have the same information, but with $p$ replaced with a directed morphism from $a$ to $a'$ and $B$ somehow functorial. Now transporting $b$ along $p$ means $B(p)(b)$, where $B(p)$ is the functorial action of $B$ on morphisms in $A$. A dependent arrow from $b$ to $b'$ over $p$ is a morphism $B(p)(b) \to b'$ in $B(a')$.
If $B$ is instead contravariant, then we have to transport $b'$ instead. That's because $B(p) : B(a') \to B(a)$ now, so we can't apply it to $b$ anymore. Thus, a dependent arrow from $b$ to $b'$ over $p$ should be a morphism $b \to B(p)(b')$.
In general, as we'll see below, $B$ might depend both covariantly and contravariantly on $a$, so we have to transport both $b$ and $b'$ to make sense of a morphism between them. Again, I've not seen a formalization of this, so forgive me if it's a bit fuzzy.
We'll denote this type of "dependent arrows over $p$" as $b \to_p b'$.
Let's see how this works with an example. Let $F, G: C \to D$ be functions (functors). If a directed morphism from $F$ to $G$ is to be analogous to a path from $F$ to $G$, one might expect such a morphism to be a pointwise morphism $\alpha_c : F(c) \to G(c)$ for each $c : C$.
What about the action of $\alpha$ on morphisms? For $f : c \to c'$ in $C$, this should be a dependent arrow $\alpha_c \to_f \alpha_{c'}$.
In this case, the type of $\alpha_c : F(c) \to G(c)$ depends on $c$ both covariantly and contravariantly. To get a dependent arrow from $\alpha_c$ to $\alpha_{c'}$ over $f$, we need to transport both sides. For $\alpha_c$, we can post-compose with $G(f)$ and for $\alpha_{c'}$, we can pre-compose with $F(f)$. Thus, a dependent path in this case is a 2-cell $G(f) \circ \alpha_c \Rightarrow \alpha_{c'} \circ F(f)$.
If we're thinking of $D$ as a 1-category, the 2-cells are just identities, so this reduces to naturality. If D is instead a 2-category, then what we get here is a lax natural transformation. If we want an oplax transformation, then what we should do is instead map $f : c \to c'$ to a dependent map $\alpha_{c'} \to_f \alpha_c$.
Let's try this out for dinatural transformations. This is where things get (at least for me) very fuzzy. We want to repeat the above, but allow $F$ and $G$ to be both covariant and contravariant at once. I think of $F$ and $G$ has "normally" only having a single variable, but then having a two-variable "extension".
In our hypothetical directed type theory, this sort of thing shows up all the time. For example, the type of $id_c : \hom_C(c, c)$ is not covariant or contravariant, but $\hom_C$ can be extended to a two-variable functor $C^{op} \times C \to Type$, which makes it both covariant and contravariant.
In any case, we can try our best. A morphism from $F$ to $G$ is again pointwise, but rather than $\alpha_{c, c'} : F(c, c') \to G(c, c')$, we want to think of $F$ and $G$ as being single-variable here, to get $\alpha_c : F(c, c) \to G(c, c)$.
The action of $\alpha$ on morphisms takes a morphism $f : c \to c'$ in $C$ to a morphism $\alpha_f : \alpha_c \to_f \alpha_c'$. As in the case of natural transformations, both sides need to be transported. The covariant transport takes $\alpha_c$ to $G(1, f) \circ \alpha_c \circ F(f, 1) : F(c', c) \to G(c, c')$. The contravariant transport takes $\alpha_{c'}$ to $G(f, 1) \circ \alpha_{c'} \circ F(1, f) : F(c', c) \to G(c, c')$. Sketching this out, we get the hexagon for dinaturality.
Thus, if we follow the same convention as with natural transformations, a lax dinatural transformation has a 2-cell $\alpha_c$ to $G(1, f) \circ \alpha_c \circ F(f, 1) \Rightarrow G(f, 1) \circ \alpha_{c'} \circ F(1, f)$ and an oplax dinatural transformation has the reverse direction. The key here is that the lax version maps $f$ to $\alpha_c \to_f \alpha_{c'}$.