Equivalent definitions of differential map Let $f:M\rightarrow N$ be a smooth map between smooth manifolds,
let $p\in M$ and $v\in T_{p}M$. Two different definitions of differential
maps on tangent space: let $\gamma$ be a smooth curve on $M$ representing
$v$ ($\gamma(0)=p,\gamma'(0)=v)$ and define $df_{p}(v)=(f\circ v)'(0)$.
Second definition: let $\mathcal{D}$ be a derivation at $p$, $g:N\rightarrow\mathbb{R}$
be a smooth function, then define $df_{p}(\mathcal{D})(g)=\mathcal{D}(g\circ f)$.
We want to show that the two definitions of $df_{p}$ coincide. 
Consider the matrix representation of the second definition of $df_{p}$
in local coordinates, this is: $$\left[
df_{p}=\frac{\partial\hat{f}^{j}}{\partial x^{i}}\right]$$
where $\hat{f}$ is the coordinate representation of $f$. 
In the first definition, let $x_{1},\dots,x_{n}$ be local coordinates
at $p$. Then, $\gamma(t)=(\gamma_{1}(t),\dots,\gamma_{n}(t))$ where
$\gamma_{i}(t)=x_{i}(\gamma(t))$. The matrix representation of $df_{p}$
is:$$\left[
df_{p}=\sum_{i=1}^{n}\frac{\partial f}{\partial x^{i}}\right]$$
I might have gotten the matrix representation of $df_{p}$ in the
first definition incorrectly, but of course if these matrices are
the same with respect to these coordinates, this means that the two
definitions are coincide. I am wondering are the steps I have done right.
 A: Recall that if $\alpha$ is a curve in $M$ and $h\colon M \to \mathbb{R}$ is smooth, then the derivation associated to $\alpha'(0)$ is given by $$\alpha'(0)h = (h\circ \alpha)'(0).$$
We will use this fact twice.
As in your notation, we let $f\colon M \to N$ be a smooth map between manifolds, $p \in M$, $v \in T_pM$.  Let $\gamma$ be a smooth curve on $M$ that represents $v$, and let $\mathcal{D}$ be the derivation at $p \in M$ that represents $v$.  Let $g\colon N \to \mathbb{R}$ be a smooth function.
In the first definition, we have $df_p(v) = (f\circ \gamma)'(0)$, which is a tangent vector on $N$, and therefore may be regarded as a derivation at $f(p) \in N$.  So, we can evaluate $df_p(v)$ at the function $g$:
$$\begin{align*}
[df_p(v)](g) & = [(f\circ \gamma)'(0)](g) \\
& = (g \circ (f\circ\gamma))'(0) \\
& = ((g\circ f)\circ \gamma)'(0) \\
& = \gamma'(0)(g\circ f) \\
& = \mathcal{D}(g\circ f) \\
& = df_p(\mathcal{D})(g)
\end{align*}
$$
This shows that your two definitions coincide.
Edit: I realize that I have not addressed your actual question, which is whether or not your reasoning is correct.  However, I think it is instructive nevertheless for you to see how to approach this problem without reference to coordinates and matrices.
