This question will fade from very specific to very generic.
Consider an $\lambda$-accessible category $\mathcal{K}$ and let's call $\text{Pres}_{\lambda} \mathcal{K}$ its $\lambda$-presentables. Is the following statement true?
Given a category with directed colimits $ \mathcal{C}$ and a functor $F: \text{Pres}_{\lambda} \mathcal{K}\to \mathcal{C}$, there is an extension $\bar{F} : \mathcal{K} \to \mathcal{C}.$
I do have an idea to prove it.
Consider objects $K, K' \in \mathcal{K}$ and a map $K \stackrel{f}{\to} K'$. I need to give a definition for $\bar{F}(K), \bar{F}(K'), \bar{F}(f)$.
This is my attempt: Choose a directed diagram of $\lambda$-presentables $K_i$ such that $K = \text{colim }K_i$ and same for a diagram $K'_j$ whose colimit will be $K'$.
The map $K_i \to K \to K'$ factors through $K_i \to K \to K'_{\bar{i}}$ because of presentability of $K_i$. We call this map $f_i$. Moreover there is a natural map $$\text{colim}F(K_{\bar i}) \stackrel{i}{\to} \text{colim}F(K_{ i})$$
I pose $$\bar{F}(K) = \text{colim}F(K_i) $$ $$\bar{F}(f) = i \circ \text{colim}(F(f_i)).$$
Probably with this definition composition will not work, but maybe one can make it work.
Now question gets more vague. Consider a category $\mathcal{K}$ and $\mathcal{G}$ the subcategory generated by a strong generator.
Can the following statement be true with some additional hypotesis?
Given a category with directed colimits $ \mathcal{C}$ and a functor $F: \mathcal{G}\to \mathcal{C}$, there is an extension $\bar{F} : \mathcal{K} \to \mathcal{C}.$