Let $f : N \rightarrow M$ be a smooth map. Fix $q \in N$ and consider a curve $\gamma : [0,T] \rightarrow N$ such that $\gamma(0) = q$ and $\dot{\gamma}(0) = v = \frac{d}{dt}\mid_{t=0}\gamma(t)$. Now, the differential at q is a linear map $d_qf : T_qN \rightarrow T_{f(q)}M$ such that $d_qf(v) := \frac{d}{dt}\mid_{t=0}(f\circ \gamma)(t)$. This means that if we take a tangent vector (as an equivalence class of curves) its image is a tangent vector at $f(q)$ in $M$ (assume $f$ is a diffeo).
Now we introduce coordinates on $N$ via $\gamma(t) = (x_1(t),\dots,x_n(t))$ and on $M$ as $(f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n))$. Next, fix $i \in \{1,\dots,m\}$ so that $$\frac{d}{dt}\mid_{t=0}(f_i \circ \gamma)(t) = \sum_{j=1}^n\frac{\partial f_i}{\partial x_j}(q)\frac{\partial x_j}{\partial t}\mid_{t=0}=\sum_{j=1}^n\frac{\partial f_i}{\partial x_j}(q)\dot{x}_j(0).$$Right? We just use the chain rule in here. But now if we do this with all function $f$ we get $$d_qf(v) = \sum_{i=1}^m\sum_{j=1}^n\frac{\partial f_i}{\partial x_j}(q)\dot{x}_j(0).$$ Its just the jacobian matrix times the tangent vector at q. But this map should transform this into a tangent vector in $M$ which means that it should be written as linear combination of $\{\partial_{x_1},\dots,\partial_{x_m}\}$.
For sure there is something wrong along the linea but I don’t understand what.
Edit: I feel like there should be only one sum. Its like the differential acts coordinate wise? Maybe my confusion is how I am writing the coordinates.
No, you cannot introduce coordinates on $N$ via $\gamma:I\to N$. The curve $\gamma$ is a $1$-dimensional object, so it cannot possibly give you the information you need to parameterise an arbitrary manifold $N$, even locally.
Likewise, you cannot introduce local coordinates on $M$ via the map $f:N\to M$.
Here is what you should be doing instead. You take local coordinates $(x_1,\dots,x_n)$ on $N$, and write $\gamma(t)=(\gamma_1(t),\dots,\gamma_n(t))$. You take local coordinates $(y_1,\dots,y_m)$ on $M$, and write $f(x_1,\dots,x_n)=(f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n))$. Now, it is possible to express $f\circ\gamma:I\to M$ as $(f_1(\gamma_1(t),\dots,\gamma_n(t)),\dots,f_m(\gamma_1(t),\dots,\gamma_n(t))$. This allows you to differentiate $$\frac{d}{dt}\Big|_{t=0}(f\circ\gamma)(t)=\begin{pmatrix}\frac{d (f_1\circ\gamma)}{dt}\\\dots\\\frac{d(f_n\circ\gamma)}{dt}\end{pmatrix}=\sum\frac{d(f_i\circ\gamma)}{dt}\partial_{y_i}=df_{x_0}(v)$$ where $x_0=\gamma(0)$ and $v=\dot{\gamma}(0)\in T_{x_0}N$.