The important point to see the relation is that a matrix does not do arbitrary transformations on the vector, but linear transformations. Using this fact, the two viewpoints are easily mapped to each other.
For this behalf, let's give the two views of the matrix their own name, so we can more easily talk about them. The 3Blue1Brown view (respectively its generalization to arbitrary dimensions) is of the matrix as transformation
$$\mu:\mathbb R^n\to\mathbb R^m, v\mapsto Mv$$
where $M$ is an $m\times n$ matrix (that is, $m$ rows, $n$ columns). That is, the transformation takes an input vector $v$ and turns it into an output vector $\mu(v)$. I'll call that the transformation viewpoint.
On the other hand, Gilbert Strang's view is of the matrix as a list of $n$ column vectors $c_j\in\mathbb R^m$, and the vector $v$ as a list of $n$ numbers $v_j$ (arranged in a column) that tells us how to combine the column vectors. I'll call that the column viewpoint.
So now let's start from the transformation viewpoint, and first remember that the standard basis in $\mathbb R^n$ consists of vectors $e_j$ which have an $1$ in row $j$ and a $0$ everywhere else, or short, $(e_j)_k=\delta_{jk}$
Now we can write each vector $v$ in this basis as linear combination $$\sum_{j=1}v_je_j$$
where due to the special form of the basis, the $v_j$ are exactly the entries of the column vector.
Next, recall the fact mentioned in the beginning, that the transformation $\mu$ is not arbitrary, but a linear transformation. That is, with scalars $a,b$ and vectors $u,v$ we have
$$\mu(a u+b v)=a \mu(u)+b \mu(v).$$
Now let's apply this to the expansion of $v$ in the vector basis:
$$\mu(v)=\mu(\sum_{i=1}^n v_je_j) = \sum_{i=1}^n v_j\,\mu(e_j)$$
Thus the result of the transformation is the linear combination of the $\mu(e_j)$ given by the vector components $v_j$.
But what are those $\mu(e_j)$? Well, rember the definition of $\mu$:
$$(\mu(e_j))_l = (Me_j)_l = \sum_k M_{lk}\delta_{jk} = M_{lj} = (c_j)_l$$
In other words, the image of $e_j$ is simply $c_j$, the $j$-th column vector of M!
So in summary we get
$$\mu(v) = \sum_{j=1}^n v_j c_j$$
where on the left you have the transformation viewpoint, and on the right you have the column viewpoint. Thus this equation tells you that both viewpoints are ultimately equivalent.
On the way, we have also seen that the column vectors are nothing else than the transformed standard basis vectors.
Now both views have their strength and weaknesses. The transformation view has the strength of being basis independent. In that sense it is the more geometric view, as the basis is something that is not inherent in the geometry, but a choice by you (the fact that you can choose a basis is of course inherent, as are certain properties of it, but the basis is not fixed by those properties). On the other hand, it de-emphasizes the linearity.
The column viewpoint, on the other hand, stresses the linearity, as it explicitly looks at the linear combinations of columns. On the other hand, it is basis dependent, and therefore de-emphasizes the geometric aspects of linear transformations.
