Well, I decided to include some additional information.
Definition 1. Let $\mathcal{C}$ and $D$ be categories, $T\colon\mathcal{C}\to\mathcal{D}$ and $S\colon\mathcal{D}\to\mathcal{C}$ be functors. Then the pair $(T,S)$ is called an equivalence iff $S\circ T\cong I_{\mathcal{C}}$ and $T\circ S\cong I_{\mathcal{D}}$. In this case functors $T$ and $S$ are also called equivalences, and categories $\mathcal{C}$ and $\mathcal{D}$ are called equivalent.
It is a basic definition and Tim's answer shows why we need to use equivalence even more frequently than isomorphism. Here's another important definition:
Definition 2. Let $\mathcal{C}$ be a category, $\mathcal{S}$ be a subcategory of $\mathcal{C}$. Then the category $\mathcal{S}$ is called a skeleton of $\mathcal{C}$ iff it is a full subcategory of $\mathcal{C}$ and every object of $\mathcal{C}$ is isomorphic to precisely one object of $\mathcal{S}$.
Note, that if the axiom of choice holds, then every category has a skeleton. See also nLab article. The connection between equivalences of categories and their skeletons is described in the following proposition:
Proposition 1. Let $\mathcal{C}$ and $\mathcal{D}$ be categories, $\mathcal{S}_{\mathcal{C}}$ and $\mathcal{S}_{\mathcal{D}}$ be their skeletons. Then $\mathcal{C}\simeq \mathcal{D}$ iff $\mathcal{S}_{\mathcal{C}}\cong\mathcal{S}_{\mathcal{D}}$.
The proof follows from the fact that every category is equivalent to its skeleton and if two skeletal categories are equivalent, then they are isomorphic. You can also search Mac Lane's "Categories for the working mathematician" for the details.
Thus we can use the notion of skeleton instead of the original definition of equivalence, but sometimes it is not a simplification. As it was mentioned, even an attempt to prove that a category has a skeleton may lead to the set-theoretical difficulties. Tim also gave arguments.
You write: But why go through this whole process of defining this notion of equivalent categories if we can just create a single equivalence classes of objects via the equivalence relation of being isomorphic, and make morphisms defined on the same equivalence class "the same?"
Okay, it could be a good idea if we want to define something like a skeleton. But it isn't, because the straightforward applying this idea leads to wrong definition. Let's try to do this.
Definition 3. Let $\mathcal{C}$ be a category. Then define the graph $\text{Equiv}(\mathcal{C})$ in the following way: $\text{Obj}(\text{Equiv}(C))=\text{Obj}(\mathcal{C})/\cong_{\mathcal{C}}$ and $$\text{hom}_{\text{Equiv}(\mathcal{C})}([a],[b])=(\coprod_{a'\in[a],b'\in[b]}\text{hom}_{\mathcal{C}}(a',b'))/[(f\sim g)\Leftrightarrow(\exists a,b\in \text{Iso}(\mathcal{C})|\quad g\circ a=b\circ f) ].$$
But the graph $\text{Equiv}(\mathcal{C})$ doesn't inherit the composition law from $\mathcal{C}$. The graph $\text{Equiv}(\mathcal{C})$ doesn't even coincide with graph of any skeleton of $\mathcal{C}$ in general case. For example, it may paste two morphisms with the same domain: in the category $\mathbf{Finord}$ we have $\text{end}_{\text{Equiv}(\mathbf{Finord})}([2])=\text{hom}_{\text{Equiv}(\mathbf{Finord})}([2],[2])\cong2$, but $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.
Best Answer
Equivalent categories are identical except that they might have different numbers of isomorphic "copies" of the same objects. One way of making this precise is as follows. Say a category $\mathcal{C}$ is skeletal if isomorphic objects of $\mathcal{C}$ are equal. Given any category $\mathcal{C}$, you can find an equivalent skeletal full subcategory (or "skeleton") $\mathcal{D}$ of $\mathcal{C}$ by just taking one representative of each isomorphism class of objects (the inclusion functor $\mathcal{D}\to\mathcal{C}$ is then an equivalence). Furthermore, if $\mathcal{C}$ and $\mathcal{D}$ are both skeletal, and $F:\mathcal{C}\to\mathcal{D}$ is an equivalence, then $F$ is actually an isomorphism (this is easy to see from the characterization of equivalences as fully faithful and essentially surjective; note, however, that an inverse of $F$ as an equivalence might not be an inverse of $F$ as an isomorphism). It follows that the skeleton of a category is unique up to isomorphism, and two categories are equivalent iff their skeleta are isomorphic.
Another way to make it precise is the following. Let $F:\mathcal{C}\to\mathcal{D}$ be an equivalence of categories. Then $F$ is naturally isomorphic to an isomorphism of categories $F':\mathcal{C}\to\mathcal{D}$ iff for each object $C$ in $\mathcal{C}$, the set of objects isomorphic to $C$ has the same cardinality as the set of objects isomorphic to $F(C)$. (Proof sketch: Fix representatives of each isomorphism class in $\mathcal{C}$ and $\mathcal{D}$, together with isomorphisms from each object to the representative and bijections between the corresponding classes in $\mathcal{C}$ and $\mathcal{D}$. Then modify $F$ so that it stays the same on the representatives, but on all other objects it is given by what it does on the representatives together with the data above.) So in this sense, if you know a category up to equivalence, the only thing you don't know about it is how many isomorphic copies of each object it has.
More importantly, pretty much every useful thing you can say about objects in a category is invariant under isomorphisms. So, you never care about the distinction between an object and some other object that is isomorphic to it (at least, when you have chosen a specific isomorphism between them). In the case of locally free sheaves and vector bundles, the point is not that locally free sheaves and vector bundles are literally in bijection with each other; rather, the point is that isomorphism classes of locally free sheaves are in bijection with isomorphism classes of vector bundles, in a way compatible with the maps between these objects. When you unravel what this compatibility should mean, you find that an equivalence of categories exactly gives you the compatibility you are looking for.