Piecewise smooth vector field along a one-parameter family of curves

Let $$(M,g)$$ be a Riemannian manifold. According to Lee's book on Riemannian manifolds, a one-parameter family of curves is defined as a continuous map $$\Gamma:J\times I\to M$$, where $$I,J$$ are intervals on the real line. This map got its name since we can thus obtain two collection of curves in $$M$$: the main curves $$\Gamma_s(t)=\Gamma(s,t)$$ defined by holding $$s$$ constant, and the transverse curves $$\Gamma^{(t)}(s)=\Gamma(s,t)$$ defined by holding $$t$$ constant.

Now we can introduce the notion of a vector field along such family of curves. A vector field along $$\Gamma$$ is a continuous map $$V:J\times I\to TM$$ such that each $$(s,t)\in J\times I$$ is assigned a vector $$V(s,t)\in T_{\Gamma(s,t)}M$$. One example of such is the velocity vector field of a transverse curve, denoted by $$(\partial_s\Gamma)(s_0,t_0)={\Gamma^{(t_0)}}'(s_0).$$

I'm sorry that still another definition needs to be introduced. $$\Gamma$$ is said to be admissible if:

(i) The domain of $$\Gamma$$ is of the form $$J\times[a,b]$$ for some open interval $$J$$.

(ii) There is a partition $$(a_0,\ldots,a_k)$$ of $$[a,b]$$ such that $$\Gamma$$ is smooth on each rectangle $$J\times[a_{i-1},a_i]$$. Partitions like this are called admissible.

(iii) Every main curve is a piecewise regular curve segment. A curve is said to be regular if it has non-vanishing velocity.

Given an admissible family $$\Gamma$$, we describe a continuous vector field along $$\Gamma$$ as piecewise smooth if the restriction of the vector field to each $$J\times[a_{i-1},a_i]$$ for some admissible partition $$(a_0,\ldots,a_k)$$. Lee claims that $$\partial_s\Gamma$$ is one such vector field, and his argument about continuity of $$\partial_s\Gamma$$ on the whole $$J\times[a,b]$$ is confusing me:

To see that is continuous on the whole domain $$J\times[a,b]$$, note on the one hand that for each $$i=1,\ldots,k-1$$, the values of $$\partial_s\Gamma$$ along the set $$J\times\{a_i\}$$ depend only on the values of $$\Gamma$$ on that set, since the derivative is taken only with respect to the $$s$$ variable; on the other hand, $$\partial_s\Gamma$$ is continuous (in fact smooth) on each sub-rectangle $$J\times[a_{i-1},a_i]$$ and $$J\times[a_{i},a_{i+1}]$$, so the right-hand and left-hand limits at $$t=a_i$$ must be equal.

Why is Lee concerned about the values of $$\partial_s\Gamma$$ along $$J\times\{a_i\}$$? Is he doing something like $$\lim_{x\to a}f(x)=f(a)?$$ Thank you so much for your patience. Thank you.

It might be helpful to consider also the other "vector fields" along $$\Gamma$$: $$\partial_t \Gamma (s_0, t_0) = \Gamma_{s_0}'(t_0).$$ Unlike $$\partial_s\Gamma$$, this "vector field" is not well-defined: at $$J \times \{ a_i\}$$, although $$\Gamma$$ is smooth in $$J \times [a_{i-1}, a_i]$$ and $$J\times [a_i, a_{i+1}]$$ respectively, it is not clear if $$\lim_{t\to a_i^-} \partial_t \Gamma, \ \ \lim_{t\to a_i^+} \partial _t \Gamma$$ are the same.
For a concrete example, consider $$\Gamma : \mathbb R \times [-1, 1] \to \mathbb R^2, \ \ \Gamma (s, t) =(t, s+ |t|).$$ This is a piecewise-smooth family of curves: indeed $$\Gamma$$ is smooth when restricted to $$\mathbb R\times [-1, 0]$$ and $$\mathbb R \times [0,1]$$ respectively. However, one cannot define $$\partial_t \Gamma$$ along $$\mathbb R\times \{0\}$$.
So, $$\Gamma$$ is not really differentiable. So one must justify why $$\partial _s\Gamma$$ is well-defined, and even continuous. And the main reason is, as described in the book, $$\partial_s \Gamma$$ along $$J \times \{ a_i\}$$ depends only on the values of $$\Gamma$$ on $$J \times \{a_i\}$$.
To see that $$\partial_s \Gamma$$ is continuous, they are using this simple fact: if $$F, G$$ are continuous functions defined on $$[a, b]$$, $$[b, c]$$ respectively on $$F(b) = G(b)$$, then $$H (x) = \begin{cases} F(x) & \text{ if } x\in [a, b], \\ G(x) & \text{ if } x\in [b, c].\end{cases}$$ is also continuous.