The equation you gave is indeed the definition of matrix multiplication, applied to a $d\times d$ matrix and a $d\times 1$ matrix. But the underlying concept is something more.
The thing about vectors is that they exist, in some sense, independent of the numbers used to represent them. For example, an ordinary 3D displacement vector represents a physical length and a physical direction. These things are not numbers, they are abstract ideas. You only get the numbers when you choose a coordinate system and then compare the vector to the coordinate axes. Different coordinate systems will give you different sets of numbers for the same vector.
Two coordinate systems can be related by transformations, such as rotation and reflection. In other words, given coordinate system A, you can identify some transformation that turns it into coordinate system B, and you can come up with a $d\times d$ matrix, $R_{d\times d}$, that represents that transformation. What makes a vector a vector is that the numbers that describe the vector in coordinate system A and the numbers that describe the vector in coordinate system B are related by the same matrix.
$$\begin{pmatrix}v_B^1 \\ \vdots \\ v_B^d\end{pmatrix} = R_{d\times d}\begin{pmatrix}v_A^1 \\ \vdots \\ v_A^d\end{pmatrix}\tag{1}$$
The group of all possible transformations has some name. For example, $SO(3)$ is the group of all rotations in 3D space. Accordingly, anything that behaves as a vector (i.e. it follows equation 1) when you rotate the 3D coordinate system is called a vector of $SO(3)$, or an $SO(3)$ vector.
In case this seems like it should be obvious, let me point out that there are sets of quantities which don't behave this way, especially when you start talking about other kinds of transformations besides 3D rotations. For example, all possible Lorentz transformations, including both rotations and boosts, form the group $SO(3,1)$. The energy and momentum (of a single particle) form a vector of $SO(3,1)$, because they change in accordance with equation (1) (with $R_{d\times d}$ being a Lorentz transformation matrix) when you change reference frames. But the electromagnetic field does not. You actually need two factors of $R_{d\times d}$ to account for how EM fields change between reference frames. That makes the EM field a rank-2 tensor of $SO(3,1)$.
I would also refer you to this question of mine on Math about the meaning of a "physical vector space", which touches on the difference between a mathematical vector and a physical vector. Only the latter is subject to the requirement of equation (1).