Let $n\geq 1$ and let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ function. We know that for a fixed $x=(x_1,....,x_n)$ in $\mathbb{R}^n$, the derivative of $f$ at $x$, denoted by $f'(x)$, is a continuous linear mapping : Since $$ f': \mathbb{R}^n \to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n).$$
In this particular case, we know that $\mathcal{L}(\mathbb{R}^n,\mathbb{R})\approx \mathbb{R}^{n\times n}$ so $f'(x)$ is a matrix.
Now, if we take a simple example of $g:\mathbb{R}^3\to \mathbb{R}^2$ given by $g(x,y,z) = <(a,b),f(x,y,z)>$ with $f(x,y,z)=(f_2(x,y,z),f_2(x,y,z))$. If we wanna calculate the following $$ g'(x,y,z),$$ Is this equivalent to $g'(x,y,z)= f'(x,y,z)(a,b)$ ?
I change your notation a little.
I consider a differentiable function
$$ f \;:\;\mathbb R^n \to \mathbb R^m $$
and the function
$$ g\;:\;\mathbb R^n \to \mathbb R\qquad\quad g(x) := \big(a\mid f(x)\big) $$
where $\;a\;$ is a fixed vector in $\;\mathbb R^m$, and $\;\;\big(\cdot\mid\cdot\big)\;$ is a scalar product in $\;\mathbb R^m$.
For any fixed $\;x\in\mathbb R^n\;$ and for $\;h\in\mathbb R^n\;$, we have
\begin{align} g(x+h)-g(x) &= \big(a\mid f(x+h)\big) - \big(a\mid f(x)\big)=\\[1.5ex] &= \big(a\mid f(x+h)-f(x)\big) =\\[1.5ex] &= \big(a\mid df(x)(h)+o(h)\big) =\\[1.5ex] &= \big(a\mid df(x)(h)\big) + \big(a\mid o(h)\big). \end{align}
Now, $\;\big(a\mid o(h)\big)\;$ is itself an $\;o(h)\;$ since
$$ \frac{\big|\big(a\mid o(h)\big)\big|}{\|h\|}\leq \frac{\|a\|\cdot\|o(h)\|}{\|h\|}\to 0 \qquad\text{for }\;h\to 0. $$
The preceding calculation therefore gives
$$ dg(x)(h) = \big(a\mid df(x)(h)\big) $$
or even
$$ dg(x) = \big(a\mid df(x)(\cdot)\big). $$