One-dimensional models for motion are useful when there are symmetries to exploit, which allows one to extend 1-D descriptions and solve real, 3-D systems. Vectorizing Newton's equations is necessary when modelling systems in real, 3-D space.
Transitioning our picture from $x(t) \rightarrow \vect{r}(t)$: the following will serve as a brief overview
on vector-valued functions. There are many equivalent ways of expressing vectors, and so it
is advantageous to cover a more than one when introducing the topic.
In $\vect{r}$, we have a weighted sum of three standard basis vectors which
correspond to position in the x,y,z directions:
Eq. 1 - A vector in $\Bbb{R}^3$ measured from the origin (0, 0, 0). expressed in four equivalent forms.
Since our $\vect{r}$ exists in flat space, then the magnitude of our vector is the pythagorean sum of its elements:
$$\bbox[8px,border:0px solid black]{ |\vect{r}(t)| = \sqrt{\vect{r} \cdot \vect{r}} = x(t) \oplus y(t) \oplus z(t) = \sqrt{[x(t)]^2+[y(t)]^2+[z(t)]^2}} \tag{Eq. 2} $$Notice, our radicand is not a sum of vector-valued quantities; $x, y, z$ are scalar functions of t. Operating on these scalar functions with the unit-vectors ($\hata{x}, \hata{y}, \hata{z}$) results in $\vect{r}$'s directional information.
Introducing our calculus operations: differentiating $\vect{r}$ with respect to time will simply apply the scalar operation to each component, giving us definitions for $\vect{a}$ and $\vect{v}$ that are analogous to our one-dimensional model:
$$\bbox[8px,border:0px solid black]{\vect{a}(t) = \frac{d \vect{v}}{dt} = \frac{d^2 \vect{r}}{dt^2} = \frac{d^2}{dt^2} \vecta{x(t)}{y(t)}{z(t)} = \vecta{\frac{d^2 x(t)}{dt^2}}{\frac{d^2 y(t)}{dt^2}}{\frac{d^2 z(t)}{dt^2}} } \tag{Eq. 3} $$ $$\bbox[8px,border:1px solid black]{\vect{F}(\vect{r},t) = m \ddot{\vect{r}} = \begin{bmatrix} F_x\\ F_y\\ F_z \end{bmatrix} = m \begin{bmatrix} \ddot{x}(t)\\ \ddot{y}(t)\\ \ddot{z}(t) \end{bmatrix}} \tag{Eq. 4}$$Eq. 3 - A concise form for Newton's Equation, generalized to $\Bbb{R}^3$. Here, m is taken to be a scalar quantity, but mass is most generally a 3x3, symmetric matrix M.
Force in Newton's Equation is determined by our system choice, whereas $\ddot{\vect{r}}$ (system motion) is the quantity we're interested in obtaining.
Evidently, this vector formulation is a more compact (short-hand) way of expressing our three separate differential equations. Conceptually, this serves a greater purpose than simply cleaning-up our notation:
Notice, too, that this formulation is not limited to 3-vectors. For any choice of $\Bbb{R}^k $ ($k \in \Bbb{N}$), (Eq. 1) and (Eq. 2) can be generalized as k-vector:
$$\vect{Q} =\begin{bmatrix} x_1(t)\\ x_2(t)\\ x_3(t)\\ \vdots\\ x_k(t) \end{bmatrix}\ \text{and}\ \ | \vect{Q}| = \sqrt{ \sum_{k=1}^{k} x^2_k(t)} \tag{Eq. 5}$$Eq. 5 - Generalizations of (Eq. 1) and (Eq. 2), s.t. $\text{dim} \vect{Q} = k$. Typical (Lagrange) notation is $x_k \rightarrow q_k$.
Why would we even care about generalizing a system beyond our 3-dimensional, flat coordiante system?
Let us start by defining a full differential element for $\vect{r}$ to describe arbitrarily small changes:
Eq. 6 - Differential elements for our Cartesian unit vectors are equal to 0, as they are always constant. Other coordinate systems do not produce such a clean vector operation.
Finally, the chain rule can resolve the velocity-dependence of $m \ddot{\vect{r}}$. This is needed when expressing the energy quantities we care about (such as kinetic energy). For each vector component j in $\vect{r_j}$:
$$ \begin{align} m \ddot{x_j} &= m \frac{dv_j}{dt}\\ &=m \frac{dv_j}{dx_j} \frac{dx_j}{dt}\\ m \ddot{x_j} &= \bbox[8px,border:1px solid black]{m v \frac{dv_j}{dx_j}} \end{align} \tag{Eq. 7} $$Eq. 7 - Interpreting physical meaning from this in-between quantity is tricky.
Now we are ready to express Newton's Equation as a conservation statement. Multiplying our differential $d \vect{r}$ through (Eq. 1), we can resolve the conserved quantity $E$, the total system energy:
Integrate both sides of this expression w.r.t. the separated differential elements:
$$ \begin{align*} \int_r \vect{F}(\vect{r}) \cdot d \vect{r} &= \int_{v_{0}}^{v_x} m v_x\ dv_x + \int_{v_0}^{v_y} m v_y\ dv_y \ + \int_{v_0}^{v_z} m v_z\ dv_z \end{align*} \tag{Eq. 8} $$Eq. 8 - To keep the expression neat, consider the case where our initial system position and velocity ($r_0,v_0) are zero.
We should leave this sum as a path integral of our 3-vector $F(\vect{r})$, to keep directional information in-tact. Notice that dot products take vector inputs but produce scalar outputs, resulting the scalar-value of energy (and an energy-dependence on velocity magnitudes). $$v^2=\vect{v} \cdot \vect{v}$$
Eq. 9 - Our statement of energy conservation, with surprising brevity. Notice that velocity is no longer a vector. In fact, we have mostly transformed Newton's statement back into a scalar equation.
Compare E to the term that outputs kinetic energy ($\frac{1}{2}mv^2$):
$$\blbox{[E] = [m][v]^2 = M^1L^2T^{-2} } \tag{Eq. 9.1}$$Eq. 9.1 - In S.I. units, this is $kg\ m^2 / s^2$ -- Newton-meters -- Joules.
This dimensionality asserts that system energy may vary in time, across space, and with different system mass.
Let's take a second and appreciate what exactly our conservation statement is saying.
Eq. 9.2* - Total energy of an isolated system is conserved. This topic is developed in statistical mechanics. More rigorously, this time-energy conservation relationship is a consequence of Noether's theorem.
Eq. 10 & 11 - $\vect{F}$ and $d \vect{r}$ are determined by our system force and path. Our potential takes a vector input, but it outputs a scalar quantity. Applying the del operator ($\vec{\nabla}$) to $V(\vect{r})$ gives us its gradient.
Suddenly, our focus has shifted from the observable motion of our system ($\vect{r}$) to quantites that are more abstract ($E$ and $V(\vect{r})$). We even have a linear operator ($\vec{\nabla}$) to relate $\vect{F}$ to our potential V. Of course, $\vec{\nabla}$ is a linear operator becuase $\frac{d}{dx_i}$ is linear.