Consider the following example from $\textrm{GL}_4$:
$$\begin{pmatrix} 1 & a & b & c \\ & 1 & d & e \\ & & 1 &f \\ & & & 1 \end{pmatrix}\begin{pmatrix} 1 & a' & b' & c' \\ & 1 & d' & e' \\ & & 1 &f' \\ & & & 1 \end{pmatrix} = \begin{pmatrix} 1 & a + a' & \ast & \ast \\ & 1 & d + d' & \ast \\ & & 1 &f + f' \\ & & & 1 \end{pmatrix}$$
Though the formula for the starred entries is complicated, the values of the product corresponding to the simple roots are easy to calculate.
In general, let $G$ be a split reductive group over a field $k$, with maximal split torus $T$ and Borel subgroup $B = TU$ defined over $k$. Let $\Phi^+$ be a system of positive roots for $T$ in $G$ corresponding to $B$, and let $\Delta$ be the corresponding base. For each $\alpha \in \Phi$, let $U_{\alpha}$ be the root subgroup corresponding to $\alpha$. Then the product mapping (with any choice of ordering of $\Phi^+$)
$$\phi: \prod\limits_{\alpha \in \Phi^+} U_{\alpha} \rightarrow U$$
is an isomorphism of varieties. Let $x_{\alpha}: \mathbf G_a \rightarrow U_{\alpha}$ be a given isomorphism of algebraic groups. The natural generalization of the above example would be that if $u_1 = \phi(x_{\alpha}(a_{\alpha})_{\alpha}), u_2 = \phi(x_{\alpha}(b_{\alpha})_{\alpha})$, $u_1u_2 = \phi(x_{\alpha}(c_{\alpha}))$ for $a_{\alpha}, b_{\alpha}, c_{\alpha} \in k$, then $c_{\alpha} = a_{\alpha} + b_{\alpha}$ for all $\alpha \in \Delta$. Is this true?
If the ordering on $\Phi^+$ is chosen so that $\Delta$ comes first (in some order), and then $\Phi^+ - \Delta$ (in some order) (or the other way around), then the answer is yes.
The point is that $\Phi^+ - \Delta$ is a "special" set of positive roots, so the root subgroups $U_{\alpha}$ for $\alpha \in \Phi^+ - \Delta$ directly span a subgroup $U'$ of $U$ which is normalized by the simple root subgroups $U_{\alpha} : \alpha \in \Delta$. Hence $U'$ is normal in $U$. This follows from the material in Chapter 14 of Borel, Linear Algebraic Groups. Also, it follows from 8.23 in Springer, Linear Algebraic Groups that $U'$ contains the derived group of $U$ (I think they are equal, but I'm not sure).
To make life easy, let's pretend $\Delta$ has two elements $\alpha_1, \alpha_2$ and $\Phi^+ - \Delta$ has two elements $\alpha_3, \alpha_4$ (although no root system looks like this). If $a_1, a_1', ... , a_4,a_4' \in k$, and $x_1, ... , x_4$ are the root vectors, then
$$x_1(a_1)x_2(a_2)x_3(a_3)x_4(a_4) \cdot x_1(a_1')x_2(a_2')x_3(a_3')x_4(a_4')$$
can be modified by placing $x_1(a_1')x_2(a_2')x_2(a_2')^{-1}x_1(a_1')^{-1}$ in between $x_2(a_2)$ and $x_3(a_3)$. Since $U$ normalizes $U'$, this becomes
$$x_1(a_1)x_2(a_2)x_1(a_1')x_2(a_2') \cdot \textrm{(something in U')}$$
We are done by 8.23, since we can switch the order of $x_2(a_2')$ and $x_1(a_1')$ and multiply something in $U'$ by the right, because $U'$ contains all commutators.