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Let's observe a natural isomorphism $\lambda$: $T_p \Bbb R^n \mapsto \Bbb R^n$. Then we have $\lambda(\dfrac{\partial }{\partial x_i})=e_i$. But when we write for $v \in T_p \Bbb R^n : v=v^i\dfrac{\partial }{\partial x_i}=v^ie_i$ although the equality is true only considering isomorphism $\lambda$. Why don't we write $\lambda(v)=v^ie_i$ instead? Is an isomorphism such a strong morphism, that the elements of different vector spaces are considered identical?

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I have no idea what you're asking but your description of the isomorphism is somewhat misleading. It is true that the identification of $T_p{\mathbb{R}^n}$ and $\mathbb{R}^n$ is obtained by the map you write but it makes it look like there is some choice involved. I mean, why should we send $\frac{\partial}{\partial x^i}$ to $e_i$ (the elements of the standard basis) and not to some other orthonormal (or completely arbitrary) basis $f_j$ of $\mathbb{R}^n$?

The issue is best understood for a slightly more abstract point of view. Let $V$ be a real finite dimensional vector space (and in particular, it doesn't come with a natural basis). We can think of $V$ as a smooth manifold in a natural way. Then for each $p \in V$ we have a natural isomorphism $T_p(V) \approx V$. This isomorphism is given by taking a vector $v \in V$ and letting $\alpha_v(t) := p + tv$ be the curve that "translates $v$ from the origin to p". The curve $\alpha_v$ is a smooth curve and $\alpha_v(0) = p$ so $\dot{\alpha_v}(0) \in T_p(V)$. The isomorphism $V \rightarrow T_pV$ is given by $v \mapsto \dot{\alpha_v}(0)$. Once descried in this way, we see that there is absolutely no choice involved in the isomorphisms. If we want, we can choose a basis $e_1, \dots, e_n$ of $V$, use it to define linear coordinates $x^1,\dots,x^n$ on $V$ (these coordinates correspond to the dual basis of $e_1,\dots,e_n)$ and then the isomorphism will send $e_i$ to $\frac{\partial}{\partial x^i}|_{p}$ but it is defined independently of any coordinate system.

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    Thank you very much for your answer. Can you explain why do we consider $e_i $ and $ \frac{\partial}{\partial x^i}|_{p}$ as essentialy equal and write $e_i=\frac{\partial}{\partial x^i}|_{p}$ – 2017-01-16
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    I have descried natural isomorphisms $\psi_p \colon V \rightarrow T_pV$ which allow you to identify a vector $v \in V$ with a tangent vector at any point $p \in V$. Under this isomorphism in $\mathbb{R}^n$, the standard basis vectors $e_i$ correspond to the tangent vectors $\frac{\partial}{\partial x^i}|_p$ (defined as above or as directional derivative operations on functions). Sometimes one abuses notation and write $e_i = \frac{\partial}{\partial x^i}|_{p}$ instead of the more accurate but cumbersome $\psi_p(e_i) = \frac{\partial}{\partial x^i}|_p$. – 2017-01-16
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    This is denote implicitly for example when considering vector fields on $\mathbb{R}^n$. Often in elementary treatments a vector field on $\mathbb{R}^2$ is just a vector valued function $F(x,y) = (F_1(x,y),F_2(x,y))$. Using the identification we identify $F(x,y)$ with a true vector field $(x,y) \mapsto \psi_p(F(x,y)) = F_1(x,y)\frac{\partial}{\partial x^1}|_{F(x,y)} + F_2(x,y) \frac{\partial}{\partial x^2}|_{F(x,y)}$ in the sense of differential geometry. Often, this is just written as $F(x,y) = F_1(x,y)\frac{\partial}{\partial x^1}|_{F(x,y)} + F_2(x,y) \frac{\partial}{\partial x^2}|_{F(x,y)}$ – 2017-01-16
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    but this is a useful abuse of notation implying implicitly the identifications $\psi_p$. – 2017-01-16