This area of math -- covering spaces, Lie groups, representations, etc, is often presented abstractly because there are some very powerful and beautiful theorems that, to a mathematician, really clarify what is happening. But to an engineer, it is a hard slog unless you have some firm examples in mind, and you don't really need the powerful results to work everything out concretely. It just saves you a lot of time to do that.
Nevertheless, I think it's still a good and important idea to work things out concretely a few times and for that all you really need is linear algebra.
That said, the concrete version of your statement is as follows:
SO(3) is best defined as the group of all rotations in 3 space. You then show that this is just all 3x3 matrices that are orthogonal (their transpose is the inverse) and have determinant 1.
You can do this by abstract linearity arguments (e.g. the rotation of a vector times a scalar is the scalar times the rotation of the vector) or by directly writing things out with linear algebra.
The first ingredient is to realize that the rotation in the plane by angle t is a linear map of the plane to itself, and can be represented by matrix multiplication and thus a square 2x2 matrix which sends
(1, 0) to (cos(t), sin(t))
and
(0, 1) to (-sin(t), cos(t)).
Thus the matrix is
[cos(t), -sin(t)]
[sin(t), cos(t)]
this matrix clearly has determinant = 1 and you can verify that the transpose is the inverse. But you could have derived this from general principles that rotations are volume and orientation preserving.
Now a rotation in 3 space must fix some line and then is just a planar rotation for the plane perpendicular to the line. So you can pick a new basis in 3 space corresponding to the line, v, and then two orthonormal unit vectors so that the rotation is just the matrix
[1 0 0]
[0 cos(t) -sin(t)]
[0 sin(t), cos(t)]
for some choice of unit vector v and some angle t. Here you should realize that you need an orientation. E.g. the plane perpendicular to v is the same plane as is perpendicular to -v, but you need an orientation on the plane to figure out the direction of rotation.
Already this should tell you that SO(3) is three dimensional and you have a parametrization of (most of) SO(3) as a point on a sphere together with an angle, so it's kinda like S^2xS^1, except the parametrization breaks down when the angle is pi as you get the same rotation if you pick anti-podal directions and when the angle is zero all the points on the sphere map to the same (identity) rotation. So this parametrization is not a diffeomorphism, it's not even 1 to 1, but it is surjective, and knowing exactly how it fails to be 1 to 1 allows you to understand SO(3) completely because you can think of SO(3) as S^2xS^1 with some points identified.
All of the above relies solely the basics of linear algebra such as what you usually get in a multi-variable calculus course. You don't even need stuff like Jordan decomposition or other more advanced linear algebra topics, just the definition of linear maps, the definition of a "rotation" in 3 space, ideas of orthogonality and the determinant being an oriented volume of a linear map. Most of these concepts are taught in multi-variable calculus as you need them to get volume forms as the result of a change of basis when you are doing integrals over surfaces and volumes.
In terms of 'topological group", the set of matrices with determinant 1 that are orthogonal form a group, as is easily verified via the fact that det(A*B) = det(A)det(B) and det(A^t) = det(A). That is all you need to show that this is a group. It is a topological group in the sense that the multiplication operation is continuous in the inherited norm you expect to get on matrices. E.g. if you write out the multiplication of matrices with the entries being variables you just get polynomials in the product of the two matrices so multiplication is a continuous operation.
When you are working at the elementary level, you don't care too much about whether the matrices are topological groups because you are not going to be using the heavy duty Lie theory machinery, you can write everything out in terms of matrices and maps between them explicitly. It's really good to write things out explicitly a few times and then learn all the abstract stuff because it helps you understand what the general results are really saying. Do not be intimated by people using terms like "universal cover", homotopy, classifying spaces, etc, as you don't need any of that to understand the basic properties of quaternions and the orthogonal groups, but these abstractions have shown to be an very useful way of looking at these spaces so they can help explain what is happening in a deeper way than relying on matrix algebra once you get to the point where you are searching for unifying ideas behind these results. The results themselves can always be proved with elementary techniques.
Nevertheless, I think it's still a good and important idea to work things out concretely a few times and for that all you really need is linear algebra.
That said, the concrete version of your statement is as follows:
SO(3) is best defined as the group of all rotations in 3 space. You then show that this is just all 3x3 matrices that are orthogonal (their transpose is the inverse) and have determinant 1.
You can do this by abstract linearity arguments (e.g. the rotation of a vector times a scalar is the scalar times the rotation of the vector) or by directly writing things out with linear algebra.
The first ingredient is to realize that the rotation in the plane by angle t is a linear map of the plane to itself, and can be represented by matrix multiplication and thus a square 2x2 matrix which sends
(1, 0) to (cos(t), sin(t))
and
(0, 1) to (-sin(t), cos(t)).
Thus the matrix is
[cos(t), -sin(t)]
[sin(t), cos(t)]
this matrix clearly has determinant = 1 and you can verify that the transpose is the inverse. But you could have derived this from general principles that rotations are volume and orientation preserving.
Now a rotation in 3 space must fix some line and then is just a planar rotation for the plane perpendicular to the line. So you can pick a new basis in 3 space corresponding to the line, v, and then two orthonormal unit vectors so that the rotation is just the matrix
[1 0 0]
[0 cos(t) -sin(t)]
[0 sin(t), cos(t)]
for some choice of unit vector v and some angle t. Here you should realize that you need an orientation. E.g. the plane perpendicular to v is the same plane as is perpendicular to -v, but you need an orientation on the plane to figure out the direction of rotation.
Already this should tell you that SO(3) is three dimensional and you have a parametrization of (most of) SO(3) as a point on a sphere together with an angle, so it's kinda like S^2xS^1, except the parametrization breaks down when the angle is pi as you get the same rotation if you pick anti-podal directions and when the angle is zero all the points on the sphere map to the same (identity) rotation. So this parametrization is not a diffeomorphism, it's not even 1 to 1, but it is surjective, and knowing exactly how it fails to be 1 to 1 allows you to understand SO(3) completely because you can think of SO(3) as S^2xS^1 with some points identified.
All of the above relies solely the basics of linear algebra such as what you usually get in a multi-variable calculus course. You don't even need stuff like Jordan decomposition or other more advanced linear algebra topics, just the definition of linear maps, the definition of a "rotation" in 3 space, ideas of orthogonality and the determinant being an oriented volume of a linear map. Most of these concepts are taught in multi-variable calculus as you need them to get volume forms as the result of a change of basis when you are doing integrals over surfaces and volumes.
In terms of 'topological group", the set of matrices with determinant 1 that are orthogonal form a group, as is easily verified via the fact that det(A*B) = det(A)det(B) and det(A^t) = det(A). That is all you need to show that this is a group. It is a topological group in the sense that the multiplication operation is continuous in the inherited norm you expect to get on matrices. E.g. if you write out the multiplication of matrices with the entries being variables you just get polynomials in the product of the two matrices so multiplication is a continuous operation.
When you are working at the elementary level, you don't care too much about whether the matrices are topological groups because you are not going to be using the heavy duty Lie theory machinery, you can write everything out in terms of matrices and maps between them explicitly. It's really good to write things out explicitly a few times and then learn all the abstract stuff because it helps you understand what the general results are really saying. Do not be intimated by people using terms like "universal cover", homotopy, classifying spaces, etc, as you don't need any of that to understand the basic properties of quaternions and the orthogonal groups, but these abstractions have shown to be an very useful way of looking at these spaces so they can help explain what is happening in a deeper way than relying on matrix algebra once you get to the point where you are searching for unifying ideas behind these results. The results themselves can always be proved with elementary techniques.