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Let $C^infty_p(M)$ be the set of all germs at point $p in M$.

A germ at point $x$ is $[f]_x={gin C^{infty}(mathcal{U}_x):exists_{mathcal{O_x}subsetmathcal{U}_x} g_{|_mathcal{O_x}}=f}$, where $mathcal{U}_x$ and $mathcal{O_x}$ are open neighbourhoods of $x$, in $M$.

Why do we always have $C^infty_pneq C^infty_q$ when $pneq q$?

I need to understand this in order to understand how the tangent spaces 'as the' set of derivations are always disjoint for different points.

A smooth germ at $p$ is an equivalence class of pairs $(U,f)$ consisting of an open neighbourhood $U$ of $p$ and a smooth function $f$ on $U$, modulo the equivalence relation that $(U,f) sim (V,g)$ if there is an open neighbourhood $W$ of $p$, $W subset U cap V$, such that $f|_W = g|_W$. We denote the collection of all smooth germs at $p$ by $C_p^infty$.

If $p neq q$, no germ at $q$ can belong to the set of germs at $p$. For, suppose $f in C_q^infty cap C_p^infty$. Let $(U_1,g_1)$ be a representative of the equivalence class of $f$ in $C_p^infty$, and let $(U_2,g_2)$ be a representative of the equivalence class of $f$ in $C_q^infty$. We can always choose the open neighbourhoods $U_1$ and $U_2$ (of $p$ and $q$, respectively) in such a manner that $U_1 cap U_2 = emptyset$, since $Bbb{R}^n$ is Hausdorff.
Now, since $g_1$ and $g_2$ are both representatives of $f$, it must be that they agree on an open set $V subset U_1 cap U_2$, where $V$ is an open neighbourhood of both $p$ and $q$. But, this is not possible since $U_1 cap U_2 = emptyset$.

Hence, $C_p^infty$ and $C_q^infty$ are completely distinct when $p neq q$.

Edit (based on comments for further clarification):

I was rereading your answer, and I have a doubt. In the 2nd paragraph, why is there such a $V$? The representatives of f are representatives, but at different points... ;)

Brahadeesh, I've added the definition I'm using of germ. With that definition, I fail to see how we can guarantee the existence of such a V.

The same (or similar) reasoning goes through with the definition of germ that you updated in the question details. Suppose $[f_x]=[f_y]$ for some $x neq y$, as per your definition. Then, they have the same set of representatives. Let $U_x$ and $U_y$ be disjoint open neighbourhoods of $x$ and $y$, respectively. Choose $g in C^infty(U_x)$ to be a representative of $[f_x]$, so that there exists $O_x subset U_x$ with $g|_{O_x} = f$. Since $[f_x]=[f_y]$, $g$ is also a representative of $[f_y]$, which is absurd.

At this point, it depends on what you want to say: it is absurd because $y notin U_x$ itself, so you are done. But if it’s not immediately clear with this, you could proceed a little bit ahead with the absurdity and observe it happen at another point, as I did previously in my answer. So, let $h in C^infty(U_y)$ be a representative of $[f_y]$. Since $[f_x]=[f_y]$, we now have two representatives of $[f_x]$. Hence there is a third representative $k in C^infty(U)$ of $[f_x]$ such that $U subset U_x cap U_y$. But this is absurd because $U_x cap U_y = emptyset$.

Also, if your reasoning was correct, then how come, in Arthur's answer, we can find a common function at germs in different points?

A germ at a point is not just a smooth function with (some conditions), it is an equivalence class of pairs of smooth functions and neighbourhoods of that point satisfying (some conditions). Arthur does not say in his answer that we can find a common germ, so there is no contradiction.

They are isomorphic as $Bbb R$ vector spaces (or algebras, for that matter), but completely distinct. They are both partitions of $C^infty(Bbb R)$ into equivalence classes (or something closely analogous, depending on your exact definition of "germ"), but those two partitions have little to do with one another.

For instance, given any two germs $alphain C^infty_p(Bbb R^n)$ and $betain C^infty_q(Bbb R^n)$, there is an $fin C^infty(Bbb R^n)$ which is contained in both $alpha$ and $beta$.