The following is a well known fact and due to the functorial properties of the jet functor:

Suppose you have two smooth manifolds $M$ and $N$ and maps $f:M \rightarrow N$ as well as $g: M \rightarrow N$, then if there is a $m \in M$ and an open neighbourhood $U$ of $m$ in $M$ such that the equation

$$f(x) = g(x)$$

holds on $U$ then clearly $j^1_mf = j^1_mg$.

The inverse Problem is when you have jet equivalence classes $j^1_mf, j^1_mg \in J^1_m(M,N)$ with $$j_mf = j_mg$$

then is there an open neighbourhood $U$ of $m$ in $M$ and representatives $f \in j_mf$ as well as $g \in j_mg$ such that the equation $f(x) = g(x)$ holds on $U$ ?

Surely there are, but what is the argument in situations like this? (The functor properties of $J^1$ doesn't really fits here I guess, because the problem is inverse)

The point is, that I can't come up with a formal correct argument and when the situation gets more involved it is hard to say if the inverse question has a solution or not, due to the lack of an correct argument.

For example:

Suppose you have smooth manifolds and jet equivalence classes

$$j_{m_1}^1f \in J^1_{m_1}(M_1,N_1)$$ $$j_{m_2}^1h \in J^1_{m_2}(M_2,N_2)$$ $$j_{n_1}^1g_1, j_{n_1}^1g_2 \in J^1_{n_1}(N_1,N_2)$$ $$j_{m_1}^1k_1,j_{m_1}^1k_2 \in J^1_{m_1}(M_1,M_2)$$

(with $k_1(m_1) = m_2 = k_2(m_1)$, $f(m_1)=n_1$, $g_1(n_1)=h(m_2)=g_2(n_2)$ and a system of two jet equation of first order jets

$$j^1_{n_1}g_1 \circ j^1_{m_1}f = j^1_{m_2} h \circ j^1_{m_1}k_1$$ $$j^1_{n_1}g_2 \circ j^1_{m_1}f = j^1_{m_2} h \circ j^1_{m_1}k_2$$

and you know that this equation holds! So the question is not whether or not the jets satisfy the equation, but the equation is taken for granted.

But the inverse problem is the question whether or not there are appropriate reprensentatives $g_1' \in j^1_{n_1}$ $\ldots$ (and all others) and an open neighbourhood $U$ of $m_1$ in $M_1$ such that

$$g_1 \circ f(x) = h \circ k_1(x)$$ $$g_2 \circ f(x) = h \circ k_2(x)$$

holds on $U$.

THis last example is not obvious and a formal argument is really needed.