I learned that the derivative (or pushforward) of a smooth map $f:M\rightarrow N$ is a map linear $f_*(p)$ between tangent spaces such that $f_*(p)([\gamma]_p)=[f\circ \gamma]_{f(p)}$. I have also learned that we can define the tangent space elements to be derivations of $\mathcal{C}^\infty(U)$ where $U\subseteq M$.
If we now consider the case $M=\mathbb{R}^m$, a basis of $T_x\mathbb{R}^m$ is $\{\frac{\partial}{\partial x_j}(x)=\partial_j(x)\}$.
For $f:\mathbb{R}^m\supseteq U\rightarrow\mathbb{R}$, I have seen the following: $$f_*(\partial_i(x))=\partial_if(x)$$
But I do not know why this is true: first of all, I think that there is an $x$ implicit: $f_*(\partial_i(x))=f_*(x)(\partial_i(x))$. Now, I only know how to apply the derivative to classes of curves, not to derivations. Even if I identify $\partial_i(x)$ with $[\delta_i]_x$ (element of the standard basis for $T_x\mathbb{R}^m$ when seen has a set of equivalent classes of curves), I get $$f_*(x)([\delta_i]_x)=[f\circ \delta_i]_{f(x)}$$ My only guess now it to do the following:I want to see to what derivation $v_x$ does $[f\circ \delta_i]_{f(x)}$ correspond. This can be accomplished by considering $$v_x:\mathcal{C}^\infty(U)\rightarrow \mathbb{R} \\ g\mapsto (g\circ(f\circ\delta_i))'(0)$$ and $$ (g\circ(f\circ\delta_i))'(0)=Dg(f(\delta_i(0))Df(\delta_i(0)\delta'_i(0)=Dg(f(x)Df(x)(e_i)$$ But this is not hte same as $\partial_i f(x)(g)$.
You have to understand that $f_{*}$ is just a generalization of the linear map $Df(p)$ i.e the original differential of a smooth function $f: \mathbb{R}^n \to \mathbb{R}^m$. To answer your question, recall that if $X_p \in T_pM$ and $f: M \to N$ smooth with $g \in C^{\infty} (V \cap f(U))$ then we define,
$$f_{*} X_p (g) = X_p(g \circ f) \in \mathbb{R}$$
i.e under $f_*$ we have $X_p \mapsto f_* X_p$. You can easily show that if $\gamma$ is the curve such that $\gamma'(0) = X_p$ then $f_* X_p = (f \circ \gamma)'(0)$ and here you will see what I mean about this just being a generalization of the usual definition. It serves as a means to prove theorems and works well in doing so since it is a coordinate free definition.
Now, let $(U, \phi) = (U, x^1,...,x^n)$ be a chart on $M$ which contains $p$. Then as you remarked, $\partial ^i(p)$ forms as basis for $T_pM$ where $i = 1,...,n$. Similarly, about $f(p):=q$ we have $\partial/\partial y^j$ is a basis for $T_qN$. We now come to your problem, I will proceed without of the messy evaluations.
\begin{align*}f_* \partial^i(p) \in T_{f(p)}N &\Rightarrow f_* \frac{\partial}{\partial x^i} = \sum_{j=1}^n a^i_j \frac{\partial}{\partial y^j} \\ & \Rightarrow f_*\frac{\partial}{\partial x^i} (y^j) =\sum_{j=1}^n a^i_j \frac{\partial}{\partial y^j} (y^j) \\ & \Rightarrow \frac{\partial (y^j \circ f)}{\partial x^i} = \sum_{j=1}^n a^i_j \delta^i_j = a^j_j \end{align*}
Now if we drop $y^j$ from the expression, we have:
$$f_* \frac{\partial}{\partial x^i} = \frac{\partial}{\partial x^i} (f)$$
The key here too is to observe that dropping the $y^j$ evaluation truly is valid since by definition,
$$ \frac{\partial}{\partial x^i} f \ (y^j) = f_* \frac{\partial}{\partial x^i} (y^j) = \frac{\partial (y^j \circ f)}{\partial x^i}$$
What should be appreciated more is the geometry of this statement which is, $\partial f/ \partial x^i$ can be thought of as the $x^i$ coordinate direction in the standard Euclidean plane.
$\textbf{Clarification}$:
$$ f_* \frac{\partial}{\partial x^i} (y^j) :=\frac{\partial}{\partial x^i}(y^j \circ f)= \frac{\partial (y^j \circ f)}{\partial x^i}$$
now drop the $y^j$.