Intersection of transversal curves Exercise $1.17$ in Montiel & Ros' Curves and Surfaces states

Two regular plane curves $\alpha$ and $\beta$ defined, respectively, on two intervals containing the origin, intersect transversely at $p = \alpha(0)=\beta(0)$ if $\alpha'(0)$ and $\beta'(0)$ are not parallel. Show that, if this happens, the traces of $\alpha_t$ and $\beta$ have non-empty intersection, for $t$ sufficiently small, where $\alpha_t$ is the curve defined by $\alpha_t(u) = \alpha(u) + ta$, where $a \in \Bbb R^2$, $\|a\|=1$.

Later they give this sketch: if $I$ is an interval where both curves are defined, put $F\colon I \times \Bbb R \to \Bbb R^2$, $F(t,u) = \alpha_t(u) - \beta(u)$. Then $F(0,0) = 0$ and $F_u(0,0) \neq 0$, so the Implicit Function Theorem gives an interval $0 \in K \subseteq I$ and a smooth function $f$ such that $F(t,f(t)) = 0$ for all $t \in K$, that is, $\alpha_t(f(t))=\beta(t)$ for all $t \in K$.
Question: I don't understand how the IFT applies here. As far as I understand, we apply the theorem for functions of the form $F\colon \Bbb R^n \times \Bbb R^{\color{red}{k}} \to \Bbb R^{\color{red}{k}}$ where the partial derivative $D_yF(x_0,y_0)$ is non-singular. We factor a piece of the domain with the same dimension of the codomain. This is not the case here. Also, they don't use in any place the fact that $\|a\|=1$ (although it seems to me that this is a silly detail), and they only use that $\alpha'(0)$ and $\beta'(0)$ are distinct (but they still could be parallel for this reasoning). Using $\|\alpha_t(u)-\beta(u)\|^2$ instead of that $F$ does not work (the partial derivatives are zero at $(0,0)$). Is there any way to tweak the IFT so it applies? If not, how can we solve the exercise? It is easy to see that the curves really must be in $\Bbb R^2$ for the result to be true, otherwise taking $a$ to be orthogonal to both $\alpha'(0)$ and $\beta'(0)$ gives a counter-example, and a possible generalization would deal with hypersurfaces, I think.
 A: Let $ \ F: I \times I \times \mathbb{R} \to \mathbb{R}^2 \ $ be such that $$F(t,s,u) = \alpha_t(s) - \beta(u) = \alpha(s) + t \cdot a - \beta(u) \ \ . $$ Then $F$ is smooth and$$J_F (t,s,u) = \left\{ \begin{array}{ccc} a^1 & [\alpha'(s)]^1 & [\beta'(u)]^1 \\ a^2 & [\alpha'(s)]^2 & [\beta'(u)]^2 \end{array} \right\} = \big[ a \qquad \tilde{J}_F (t,s,u) \big]_{2 \times 3} \ \ \ , $$ where $ \ a = (a^1,a^2) \in \mathbb{R}^2 \ $ and $ \ \tilde{J}_F(t,s,u) = \big[ \alpha'(s) \ \ \ \ \beta'(u) \big]_{} \in \mathbb{R}^{2 \times 2}$, with $ \ \det \big( \tilde{J}_F (0,0,0) \big) \neq 0 \, $, as $ \ \{ \alpha'(0) ,\beta'(0) \} \ $ is linearly independent.
By IFT, there exists an interval $K$, with $ \ 0 \in K \subset I \, $, an open set $ \ V \subset I \times \mathbb{R} \subset \mathbb{R}^2 \ $ and a smooth function $ \ \gamma = (\gamma^1 , \gamma^2):K \to V \ $ such that $ \ F \big( t , \gamma(t) \big) = (0,0) \, $, $\forall t \in K$. That is, $\forall t \in K$, $$\alpha_t \big( \gamma^1 (t) \big) = \beta \big( \gamma^2 (t) \big) \ \ . $$
