I think your question should be matched with an answer that is equally free flowing and open minded as the question itself. So, here they are my two analogies.
First, unless you're a pure mathematician, you were probably taught univariate probabilities and statistics first. For instance, most likely your first OLS example was probably on a model like this:
$$y_i=a+bx_i+e_i$$
Most likely, you went through deriving the estimates through actually minimizing the sum of least squares:
$$TSS=\sum_i(y_i-\bar a-\bar b x_i)^2$$
Then you write the FOCs for parameters and get the solution:
$$\frac{\partial TTS}{\partial \bar a}=0$$
Then later you're told that there's an easier way of doing this with vector (matrix) notation:
$$y=Xb+e$$
and the TTS becomes:
$$TTS=(y-X\bar b)'(y-X\bar b)$$
The FOCs are:
$$2X'(y-X\bar b)=0$$
And the solution is
$$\bar b=(X'X)^{-1}X'y$$
If you're good at linear algebra, you'll stick to the second approach once you've learned it, because it's actually easier than writing down all the sums in the first approach, especially once you get into multivariate statistics.
Hence my analogy is that moving to tensors from matrices is similar to moving from vectors to matrices: if you know tensors some things will look easier this way.
Second, where do the tensors come from? I'm not sure about the whole history of this thing, but I learned them in theoretical mechanics. Certainly, we had a course on tensors, but I didn't understand what was the deal with all these fancy ways to swap indices in that math course. It all started to make sense in the context of studying tension forces.
So, in physics they also start with a simple example of pressure defined as force per unit area, hence:
$$F=p\cdot dS$$
This means you can calculate the force vector $F$ by multiplying the pressure $p$ (scalar) by the unit of area $dS$ (normal vector). That is when we have only one infinite plane surface. In this case there's just one perpendicular force. A large balloon would be good example.
However, if you're studying tension inside materials, you are dealing with all possible directions and surfaces. In this case you have forces on any given surface pulling or pushing in all directions, not only perpendicular ones. Some surfaces are torn apart by tangential forces "sideways" etc. So, your equation becomes:
$$F=P\cdot dS$$
The force is still a vector $F$ and the surface area is still represented by its normal vector $dS$, but $P$ is a tensor now, not a scalar.
Ok, a scalar and a vector are also tensors :)
Another place where tensors show up naturally is co-variance or correlation matrices. Just think of this: how to transform one correlation matrix $C_0$ to another one $C_1$? You realize we can't just do it this way: $$C_\theta(i,j)=C_0(i,j)+ \theta(C_1(i,j)-C_0(i,j)),$$
where $\theta\in[0,1]$ because we need to keep all $C_\theta$ positive semi-definite.
So, we'd have to find the path $\delta C_\theta$ such that $C_1=C_0+\int_\theta\delta C_\theta$, where $\delta C_\theta$ is a small disturbance to a matrix. There are many different paths, and we could search for the shortest ones. That's how we get into Riemannian geometry, manifolds, and... tensors.
UPDATE: what's tensor, anyway?
@amoeba and others got into a lively discussion of the meaning of tensor and whether it's the same as an array. So, I thought an example is in order.
Say, we go to a bazaar to buy groceries, and there are two merchant dudes, $d_1$ and $d_2$. We noticed that if we pay $x_1$ dollars to $d_1$ and $x_2$ dollars to $d_2$ then $d_1$ sells us $y_1=2x_1-x_2$ pounds of apples, and $d_2$ sells us $y_2=-0.5x_1+2x_2$ oranges.
For instance, if we pay both 1 dollar, i.e. $x_1=x_2=1$, then we must get 1 pound of apples and 1.5 of oranges.
We can express this relation in the form of a matrix $P$:
2 -1
-0.5 2
Then the merchants produce this much apples and oranges if we pay them $x$ dollars:
$$y=Px$$
This works exactly like a matrix by vector multiplication.
Now, let's say instead of buying the goods from these merchants separately, we declare that there are two spending bundles we utilize. We either pay both 0.71 dollars, or we pay $d_1$ 0.71 dollars and demand 0.71 dollars from $d_2$ back.
Like in the initial case, we go to a bazaar and spend $z_1$ on the bundle one and $z_2$ on the bundle 2.
So, let's look at an example where we spend just $z_1=2$ on bundle 1. In this case, the first merchant gets $x_1=1$ dollars, and the second merchant gets the same $x_2=1$. Hence, we must get the same amounts of produce like in the example above, aren't we?
Maybe, maybe not. You noticed that $P$ matrix is not diagonal. This indicates that for some reason how much one merchant charges for his produce depends also on how much we paid the other merchant. They must get an idea of how much pay them, maybe through rumors? In this case, if we start buying in bundles they'll know for sure how much we pay each of them, because we declare our bundles to the bazaar. In this case, how do we know that the $P$ matrix should stay the same?
Maybe with full information of our payments on the market the pricing formulas would change too! This will change our matrix $P$, and there's no way to say how exactly.
This is where we enter tensors. Essentially, with tensors we say that the calculations do not change when we start trading in bundles instead of directly with each merchant. That's the constraint, that will impose transformation rules on $P$, which we'll call a tensor.
Particularly we may notice that we have an orthonormal basis $\bar d_1,\bar d_2$, where $d_i$ means a payment of 1 dollar to a merchant $i$ and nothing to the other. We may also notice that the bundles also form an orthonormal basis $\bar d_1',\bar d_2'$, which is also a simple rotation of the first basis by 45 degrees counterclockwise. It's also a PC decomposition of the first basis. hence, we are saying that switching to the bundles is simple a change of coordinates, and it should not change the calculations. Note, that this is an outside constraint that we imposed on the model. It didn't come from pure math properties of matrices.
Now, our shopping can be expressed as a vector $x=x_1 \bar d_1+x_2\bar d_2$. The vectors are tensors too, BTW. The tensor is interesting: it can be represented as $$P=\sum_{ij}p_{ij}\bar d_i\bar d_j$$, and the groceries as $y=y_1 \bar d_1+y_2 \bar d_2$. With groceries $y_i$ means pound of produce from the merchant $i$, not the dollars paid.
Now, when we changed the coordinates to bundles the tensor equation stays the same: $$y=Pz$$
That's nice, but the payment vectors are now in the different basis: $$z=z_1 \bar d_1'+z_2\bar d_2'$$, while we may keep the produce vectors in the old basis $y=y_1 \bar d_1+y_2 \bar d_2$. The tensor changes too:$$P=\sum_{ij}p_{ij}'\bar d_i'\bar d_j'$$. It's easy to derive how the tensor must be transformed, it's going to be $PA$, where the rotation matrix is defined as $\bar d'=A\bar d$. In our case it's the coefficient of the bundle.
We can work out the formulas for tensor transformation, and they'll yield the same result as in the examples with $x_1=x_2=1$ and $z_1=0.71,z_2=0$.
