We consider smooth maps $f : (M,p) \to (N,q)$ mapping a fixed point $p \in M$ to a fixed point $q \in N$. The derivative at $p$ is a linear map $df_p : T_pM \to T_qN$. We can therefore compare the derivatives of smooth maps $f, g : (M,p) \to (N,q)$. We can also do that without using tangent spaces by choosing charts $\phi : U \to U'$ on $M$ around $p$ and $\psi : V \to V'$ on $N$ around $q$. Then we get "localized derivatives" $d_{(\phi,\psi)} f_p = d(\psi \circ f \circ \phi^{-1})_{\phi(p)}$ and $d_{(\phi,\psi)} g_p = d(\psi \circ g \circ \phi^{-1})_{\phi(p)}$ in the usual sense of multivariable calculus. Clearly $df_p = dg_p$ iff $d_{(\phi,\psi)} f_p = d_{(\phi,\psi)} g_p$. Note that the last equation does not depend on the choice of $\phi, \psi$.
If $N = \mathbb R^n$ we can compare the derivatives at $p$ of any two smooth maps $f, g$ (without assuming that $f(p) = g(p)$) by taking the identity on $\mathbb R^n$ as a chart around both $f(p)$ and $g(p)$. This was used e.g. in Hitchin's definition of tangent space and tangent vectors to define the cotangent space $T^*_pM$.
I try to understand whether such a comparison is possible for more general $N$. So let us consider the case $f(p) \ne g(p)$ for a general $N$.
Naively one could use charts $\psi_f$ and $\psi_g$ around $f(p)$ and $g(p)$ and form the corresponding localized derivatives. But obviously this does not make sense because these charts can be chosen independently which prevents a comparison of the localized derivatives. It would be the same as choosing in case $f(p) = g(p)$ two charts depending on $f,g$. Also using the same chart for $f(p)$ and $g(p)$ if these points lie in a common chart region $V$ does not seem to helpful since for some choice $\psi : V \to V'$ we may have $d_{(\phi,\psi)} f_p = d_{(\phi,\psi)} g_p$ and for some other choice $d_{(\phi,\psi)} f_p \ne d_{(\phi,\psi)} g_p$. Even if the chart region $V$ is diffeomorphic to $\mathbb R^n$ (which can always be achieved if $f(p), g(p)$ lie in the same component of $N$) we have this problem.
In case $N = \mathbb R^m$ (or more generally $N \subset \mathbb R^n$ open) we can circumvent the "chart choice problem" because we have the identity as a canonical chart. It does not seem that canonical charts are available on general $N$, thus this approach does not work.
But thinking again in terms of tangent spaces we see that the derivatives at $p$ of any two smooth maps $f, g : M \to \mathbb R^n$ can be compared because we have a canonical identification $T_q\mathbb R^n = \mathbb R^n$ for all $q \in \mathbb R^n$. That is, the tangent bundle $T\mathbb R^n$ has a canonical trivialization. It is not expedient to allow arbitrary trivalizations of $T\mathbb R^n$ because this would essentially reproduce the above chart choice problem.
The existence of a canonical trivialization of $TN$ is not a unique feature of $\mathbb R^n$. Each Lie group $N$ with neutral element $e$ has a canonical isomorphism $TN \approx N \times T_eN$. This should enable us to compare $df_p$ and $dg_p$ for any two smooth $f, g : M \to N$.
Another approach could be to consider manifolds with a flat connection. This gives canonical isomorphisms between any two tangent spaces. A connection is of course an additional ingredient, but I think in some cases we have a canonical one. Flat connections exist on more manifolds than on those having a trivial tangent bundle, so this is more general approach.
Do these two approaches make sense? Do they occur in the literature?
A possible problem seems to be that such a concept is not invariant under submanifold inclusion. As an example consider $S^1 \subset \mathbb R^2$. The maps $f : \mathbb R \to S^1, f(t) = e^{it}$, and $g : \mathbb R \to S^1, g(t) = -e^{it}$, are smooth. Their derivatives at any $t \in \mathbb R$ are equal if we use the canonical (Lie group) based identifications $d\tau_q : T_qS^1 \to T_1S^1$, where $\tau_q(z) = q^{-1}z$. But the maps $if, ig$ with inclusion $i : S^1 \to \mathbb R^2$ do not have the same derivatives at $t$.
