First, the Borel-Tits paper you refer to is Homomorphismes "abstraits" de groupes algebriques simples, in Ann. of Math. 97 (1973), 499-571.

Their main concern is a connected reductive algebraic group $G$ defined over an arbitrary field $k$, but they pay special attention to the case of local fields later in the paper. In Section 6 they define the subgroup $H^+$ for any connected $k$-group $H$ to be the (normal) subgroup of $H(k)$ generated by the rational points of all $k$-split unipotent $k$-subgroups of $H$. (Here $k$-split is usually automatic, for instance when working over a perfect field.) The definition is sometimes given in a slightly different form in other papers but amounts to the same thing.

Assuming $H$ is actually generated by its connected unipotent subgroups, which often happens in the cases of interest such as semisimple groups, a finite number of such subgroups will be enough to generate the group (an elementary consequence of finite dimensionality). So the answer to your basic question is yes.

The group $G^+$ plays a natural role for Borel-Tits in the study of "abstract" homomorphisms of almost-simple algebraic groups, but it comes up more generally in older work of Tits where he conjectured that for simply connected groups it might coincide with the full group of rational points. This had already been suggested in special cases by Kneser and became known as the Kneser-Tits Conjecture. But Platonov found a counterexample, which led later to the label Kneser-Tits Problem in work of Prasad-Raghunathan and others (related to the "Whitehead group", etc.). So there is a further paper trail of interest here, much of it in English. However, the long paper by Borel-Tits is definitely in French.

First, the Borel-Tits paper you refer to is Homomorphismes "abstraits" de groupes algebriques simples, in Ann. of Math. 97 (1973), 499-571.

Their main concern is a connected reductive algebraic group $G$ defined over an arbitrary field $k$, but they pay special attention to the case of local fields later in the paper. In Section 6 they define the subgroup $H^+$ for any connected $k$-group $H$ to be the (normal) subgroup of $H(k)$ generated by the rational points of all $k$-split unipotent $k$-subgroups of $H$. (Here $k$-split is usually automatic, for instance when working over a perfect field.) The definition is sometimes given in a slightly different form in other papers but amounts to the same thing.

Assuming $H$ is actually generated by its connected unipotent subgroups, which often happens in the cases of interest such as semisimple groups, a finite number of such subgroups will be enough to generate the group (an elementary consequence of finite dimensionality). So the answer to your basic question is yes.

The group $G^+$ plays a natural role for Borel-Tits in the study of "abstract" homomorphisms of almost-simple algebraic groups, but it comes up more generally in older work of Tits where he conjectured that for simply connected groups it might coincide with the full group of rational points. This had already been suggested in special cases by Kneser and became known as the Kneser-Tits Conjecture. But Platonov found a counterexample, which led later to the label Kneser-Tits Problem in work of Prasad-Raghunathan and others (related to the "Whitehead group", etc.). So there is a further paper trail of interest here, much of it in English. However, the long paper by Borel-Tits is definitely in French.

EDIT: As Yves points out, I'm not directly answering the question asked. Probably I'm not understanding the motivation here, but the Borel-Tits subgroup is always required to be normal and in good cases is just the subgroup generated by all rational unipotent elements. Typically this comes up just for reductive groups, where the unipotent $k$-subgroups live inside the unipotent radicals of minimal $k$-parabolic subgroups. There are only finitely many conjugacy classes of the latter, so it's enough to start with finitely many unipotent groups (and take the normal closure). I've never seen any situation in which other finite choices of generating subgroups would be natural, but I suspect the answer to the (perhaps artificial) question asked is in fact yes. Is the question motivated somehow?