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Loring Tu defines $C^{\infty}_p(M)$ to be the germs of a $C^{\infty}$ function at $p$, which is the equivalence class of functions defined in a neighborhood of $p$ and agree on some possibly smaller neighborhood of $p$. Then he defines a derivation at a point $p$ of a smooth manifold $M$ to be a linear map that sends $C^{\infty}_p(M)$ to $\mathbb R$ which satisfies leibnitz law. But he does not prove that this is well-defined.

To prove the concept of derivation is well-defined, I need to pick two functions $f$, $g$ which agree on some neighborhood of $p$, but they can possibly have different domains. Now I need to show that their derivation should agree. The difficulty here is that $f$ and $g$ might have different domains, then I can not use the usual bump function argument. So what should I do?

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    You did not specify your derivation completely.There are many maps $C_p^\infty(M)\to\Bbb R$ that sastify the Leibnitz law. In particular if $D$ does it then so does $cD$ for any constant $c$. – 2017-01-22
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    Maybe I should have said 'a derivation' rather than 'the derivation'. – 2017-01-22
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    The proof is algebraic. – 2017-01-22
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    @CharlieFrohman Can you refer me to a proof – 2017-01-22
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    @Keith What do you not like about my answer? – 2017-01-23

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It isn't clear quite what you are concerned about not being well defined. As I see it there are two possibilities. The first possibility as that you are hesitant to view $C_p^\infty(M)$ as a vector space over $\mathbb{R}$. Let $f,g,h$ be smooth functions to $\mathbb{R}$ defined in some neighborhoods of $p$ and let $[\cdot]$ denote the equivalence class under the equivalence relation defining $C^\infty_p(M)$. It seems possible that you are concerned that addition may not be well defined on $C^\infty_p(M)$, i.e. if we have $[f]=[g]$ then why does $[f+h]=[f]+[h]=[g]+[h]$? To check that this sort of thing is true, restrict to a neighborhood of $p$ on which all three functions are defined, so that $f$ and $g$ restricted to the neighborhood are identical.

Perhaps what you see as a problem is that Tu writes the Liebniz rule in the following form, which makes it look like it isn't a requirement related to equivalence classes. He write that a derivation $D:C^\infty_p(M)\rightarrow \mathbb{R}$ is an $\mathbb{R}$-linear map such that $$ D(fg)=D(f)g(p)+f(p)D(g). $$

Of course, this doesn't make sense literally: $D$ takes as arguments equivalence classes, i.e. germs of functions. So, we need to interpret $f,g$ as germs of functions and check that the above statement makes sense for germs. A priori, a germ doesn't have a notion of evaluating it at $p$: it is just an equivalence class of functions. If $\alpha\in C^\infty_p(M)$ is a germ of a function at $p$ then we may define $\alpha(p)$ to be $h(p)$ for some $[h]=\alpha$. Note that this is well defined because if two functions agree with $f$ in a neighborhood of $p$ then they agree at $p$. Of course, we should also check that for two germs, that $[f][g]$ is well defined, and this may be done similarly to the above paragraph.

What Hagen is alluding to is that often a particular derivation $D$ will be defined in terms of functions, so that this $D$ is constant on different germs of functions, and hence descends to a well-defined derivation. Hence as you did not show him a particular derivation to check for being well defined, there was nothing to check.