Let's first think about how we could figure out whether $u$ was in the span of $A$. We would need to see if the vector equation $a_1v_1 + a_2v_2 + a_3v_3 = u$ had a solution $(a_1,a_2,a_3)$ in real numbers. To decide whether such a solution exists, we'd write down the corresponding augmented matrix
\v11 & \v21 & \v31 & | & \u1 \\\
\v12 & \v22 & \v32 & | & \u2 \\\
\v13 & \v23 & \v33 & | & \u3 \\\
\v14 & \v24 & \v34 & | & \u4
(here $\u1,\u2,\u3,\u4$ are the components of the vector $u$, and similarly for the $v_i$), and then reduce the coefficient matrix to row-echelon form, yielding something of the form
1 & * & * & | & * \\\
0 & 1 & * & | & * \\\
0 & 0 & 1 & | & * \\\
0 & 0 & 0 & | & c_1\u1 + c_2\u2 + c_3\u3 + c_4\u4
(where $c_1,c_2,c_3,c_4$ come up during the row operations used). This system is consistent exactly when $c_1\u1 + c_2\u2 + c_3\u3 + c_4\u4=0$. In other words, $u$ is in the span of $A$ exactly when $u$ is a solution to the single equation $c_1\u1 + c_2\u2 + c_3\u3 + c_4\u4=0$.
A similar calculation will yield real numbers $d_1,d_2,d_3,d_4$ such that $u$ is in the span of $B$ exactly when $u$ is a solution to the single equation $d_1\u1 + d_2\u2 + d_3\u3 + d_4\u4=0$. Therefore, detecting whether $u$ is in the intersection of the spans of $A$ and $B$ is equivalent to deciding whether $u$ is a solution to the pair of equations
c_1\u1 + c_2\u2 + c_3\u3 + c_4\u4&=0 \\\
d_1\u1 + d_2\u2 + d_3\u3 + d_4\u4&=0.