Your intuition is correct: terms are "somehow" elements of the objects. The only thing left is to figure out what elements in a category are. They cannot be literally elements because objects need not be sets (and more generally there may be no way to even "convert" an object to a set with a functor).
If I say "consider a mass point $p$ moving in $\mathbb{R}^3$" then you will immediately understand that $p$ is not actually an element of $\mathbb{R}^3$ but a map $\mathbb{R} \to \mathbb{R}^3$, i.e., the point is parameterized by time (which we take to be a real number). It is a general phenomenon that "entities" can be parameterized, and so they can be thought of as either variable elements or maps. For instance "temperature" may be thought of as a scalar field mapping the points of Earth's surface to real numbers, or it can be thought of as a real number which varies with position on the Earth.
In any case, this leads to the idea that the elements of an object $A$ in a category are simply all morphisms $B \to A$. When $B$ is the terminal object, we get the global (non-parameterized) elements, but in general we allow elements to be parameterized by any object, and so we call them generalized elements. One way to understand the Yoneda lemma is that an object in a category consists of its generalized elements.
You asked how to pass from a category to its internal type theory. Let me answer the opposite question first. Given a type theory, we can think of terms in context as the generalized elements of a type, while the closed terms are the global elements. But since generalized elements are the same thing as morphisms this tells us how to build a category out of a type theory: the morphisms are just terms in a context, quotiented by whatever equations the theory proves.
To go in the opposite direction, suppose we have a category $\mathcal{C}$ and we want to make a type theory out of it. As you said, we take the types to be the objects of $\mathcal{C}$. According to our view that "generalized elements are morphisms", the morphisms $f : A \to B$ ought to correspond to terms of type $B$ in context $A$. But we cannot just write $x : A \vdash f : B$ because there is no $x$ on the right-hand side. It would make more sense for $f$ to have the type $A \to B$, which is precisely what we do. For each morphism $f : A \to B$ we introduce a constant of type $A \to B$, which we also write as $f$. Then the correspondence is recovered: given $x : A$ there is a term $f\, x : B$.
If our type theory does not have arrow types then we have to do things a bit differently. For each morphisms $f : A \to B$ in $\mathcal{C}$ we add a unary term constructor $f$ and the typing rule
$$\frac{\Gamma \vdash e : A}{\Gamma \vdash f\,e : B}.$$
Note that $f\,e$ is just a syntactic form, namely application of the unary term constructor $f$ to the term $e$. It would make no sense to apply the morphims $f$ to the term $e$!
A type theory consists not only of types and terms, but also of equations. Our category gives us some equations, too: if the morphism $h : A \to C$ is the composition of $f : A \to B$ and $g : B \to C$, then we get the equation
$$x : A \vdash h \, x =_C g (f \, x).$$
There are some further equations. For instance, if $A$ and $B$ are objects and $C$ is their product, then we need to relate the type $C$ with the product type $A \times B$. That is, in the category there are projections $\pi_1 : C \to A$ and $\pi_2 : C \to B$ and a pairing operation $\langle{-}, {-}\rangle$, while in type theory there are projections $\mathtt{fst} : A \times B \to A$ and $\mathtt{snd} : A \times B \to B$ and pairing $({-},{-})$ (where $A \times B$ is the formal product type). We need equations which relate these, but let me skip the technicalities.
The $\beta$ and $\eta$ reduction rules are validated by a cartesian closed category. The rules about product and function types correspond to various parts of the adjunction between cartesian products and exponentials.
In conclusion, it may seem strange that we take all morphisms in a category and make them into unary constructors of the corresponding type theory. If the category is very big that gives us a very large type theory which is not syntactic in any sense of the word syntactic. Just free your mind and the rest will follow.
