The best way to think about the derivative is:
\begin{equation*}
\tag{$\spadesuit$}f(x + \Delta x) \approx f(x) + f'(x) \Delta x.
\end{equation*}
The approximation is good when $\Delta x$ is small. This equation expresses the fact that $f$ is "locally linear" at $x$.
How can we make sense of ($\spadesuit$) when $f:\mathbb R^n \to \mathbb R^m$?
\begin{equation*}
f(\underbrace{x}_{n \times 1} + \underbrace{\Delta x}_{n \times 1}) \approx \underbrace{f(x)}_{m \times 1} + \underbrace{f'(x)}_{?} \underbrace{\Delta x}_{n \times 1}.
\end{equation*}
It appears that $f'(x)$ should be something that, when multiplied by an $n \times 1$ column vector, returns an $m \times 1$ column vector. In other words, $f'(x)$ should be an $m \times n$ matrix.
If we prefer to think in terms of linear transformations rather than matrices, we can write
\begin{equation*}
f(x + \Delta x) \approx f(x) + Df(x) \Delta x.
\end{equation*}
Here $Df(x)$ is a linear transformation that takes $\Delta x$ as input, and returns $f'(x) \Delta x$ as output. This equation is what it means to be "locally linear" in the multivariable case.
Taking this as our starting point, it's not too hard to show that
\begin{equation*}
f'(x) =
\begin{bmatrix}
\frac{\partial f_1(x)}{\partial x_1} & \cdots & \frac{\partial f_1(x)}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial f_m(x)}{\partial x_1} & \cdots & \frac{\partial f_m(x)}{\partial x_n}
\end{bmatrix}.
\end{equation*}
(The functions $f_i$ are the component functions of $f$.)
If
\begin{equation*}
X = \begin{bmatrix} X_1 \\ \vdots \\ X_n \end{bmatrix},
\end{equation*}
then
\begin{equation*}
f'(x) X =
\begin{bmatrix}
\sum_{j=1}^n \frac{\partial f_1(x)}{\partial x_j} X_j \\
\vdots \\
\sum_{j=1}^n \frac{\partial f_m(x)}{\partial x_j} X_j
\end{bmatrix},
\end{equation*}
as you can see just by doing the matrix-vector multiplication.
This is the equation given in your question.