It might be the case that the definition of finite elements from domain theory is what you are looking for. It makes precise the idea that an element of a poset carries finite amount of information.

**Definition:** Let $(P, {\leq})$ be a poset. An element $x \in P$ is **finite** (or **compact**) if it is inaccessible by directed suprema: if $D \subseteq P$ is directed and $x \leq \sup D$ then $x \leq y$ for some $y \in D$.

This is a purely order-theoretic characterization of finiteness and is therefore immune to changes in the presentation of $P$.

Looking at the [list of forcing notions](https://en.wikipedia.org/wiki/List_of_forcing_notions), it seems that the following have the property that all elements of the forcing poset are finite:

1. [Cohen forcing](https://en.wikipedia.org/wiki/Forcing_(mathematics)#Cohen_forcing)

2. Levy collapsing of an uncountable cardinal $\lambda$ to $\omega$ (but not the collapsing of $\lambda$ to $\kappa$ when $\kappa$ is uncountable – although in this case we get the related notion of *$\kappa$-finiteness).

3. Shooting a club with countable conditions.

I have not checked all the other notions, but they generally seem to contain some elements that are non-finite.

In domain theory it usually does not matter so much that a poset consists of only finite elements, but rather that every element is the supremum of finite elements below it. This is known as *algebraicity* of the poset. I would presume that the actual definition of "forcing with finite conditions" ought to be that the complete Boolean algebra associated with the notion is algebraic. But I am not set theorist, sos omeone else should confirm or deny this guess.