The only thing hard about tensors is the refusal of most expositions to point out what they really are , and to focus entirely on the notation used for them.
At risk of my sanity I will try again to explain what they are, in a simplified case, that will no doubt evoke objections, but so what.
The simplest tensor is the derivative of a function. Let's take a real valued function f in 2 variables. At each point p it has a linear approximation df(p) which is a linear function of 2 variables, namely df/dx(p) [x-x(p)] + df/dy(p) [y-y(p)]. The coordinates of this function are the two coefficients (df/dx(p), df/dy(p)).
So notationally we have assigned to each point p, a pair of numbers, or a vector
(df/dx(p), df/dy(p)). The fact that this object is linear is used to describe it by some people as a tensor of rank one, or even a tensor of "type" (0,1).
Note that if you expand your funcrtion f about p in a TAYLOR series, then this tensor of rank one is the linear part of the Taylor series.
Now what do you think we should do to approximate f better, say by the quadratic part of its Taylor series? Then we should assign to each point the second order polynomial in the Taylor series. This would assign to each point p a second degree polynomial with coefficients for x^2, xy, yx, and y^2. Of course polynomials are commutative, so we can take the xy and yx coefficients to be equal. I.e. we assign to each point an object with 4 entries which we can abbreviate as a(i,j) for i,j = 1,2, and where in this case a(1,2) = a(2,1).
This assignment of a quadratic polynomial, i.e. a symmetric bilinear function at each point, is called a symmetric "2-tensor". Some people like the symbols better than the real object, hence they refer to the notation a(i,j) as the tensor.
Similarly for every degree, the kth degree homogeneous taylor polynomial of the function defines a rank k symmetric tensor, whose notational representative is a coefficient of form a(i1,...ik), which in our case is symmetric.
Again some people refer to these symbols a(i1,...ik), as the tensor itself.
Now when we change coordinates from x,y to u,v say, then one has a change of variables transformation, given by linear formulas in the derivatives of the u's and v's wrt the x's and y's and vice versa, that can be used to express the coefficients of the u,v terms using the coefficients of the x,y terms. there is nothing more to this than substituting a linear function into each variable of a homogeneous polynomial
i.e. these coordinate transformation rules can be worked out just from knowing the nature of the given tensor as a polynomial, but some people again take these formal transformation rules as a definition of a tensor.
So instead of saying that a symmetric kth order tensor is an assignment to each point, of a kth degree polynomial on the tangent space, they say a kth order tensor is a family of coeffifcients functions a(i1,...ik) which transform by a certain totally unmotivated rule, under change of coordinates.
In my opinion this formal discussion of tensors is responsible for their reputation for difficulty.
now the subject does get more complicated, but only in two ways:
1) we allow also non commutatibve polynomials, i.e. objects whose representing symbols a(i1,...ik) are not equal when we permute the indices. After all, some physical phenomena are non commutative, like rotations in space.
2) we also look at tensors which are kth order bilinear combinations of tangent vectors, and not kth order bilinear functions on tangent vectors, i.e. we look at objects whose transformation rules require substituting the transpose of the derivative matrix, instead of the derivative matrix itself. This let's us represent geometric objects like parallelograms by tensors, as well as tenagent directions.
Now notationally this is a very compllicated business, but that argues to me for de - emphasizing the notational aspect, instead of elevating it to the height of calling it the whole subject.
My point is to keep your head about you when learning, if you want it to occur efficiently. What I have just told you might not be learned in a year of studying some books on tensors. So try to get someone to tell you what is going on, what the the tool is good for, in each case before plunging in. Since you want physics, try to get a physicist to guide you, who knows where you want to get to, and ideally is already there herself.
good luck.
