This is a very long answer, and it's hard to tell from your question how intelligible it will be to you. Hopefully it's helpful to you.
First take (In what sense is finding a universal object solving an optimization problem?)
Let's first give a precise statement about what a universal object is. A universal object is an initial or terminal object in an appropriate auxiliary category. For example, a product $X\times Y$ in a category $C$ is terminal in the category of triples $(A,f,g)$ where $A\in C$, $f:A\to X$, $g:A\to Y$, and morphisms $a:(A,f,g)\to (A',f',g')$ are arrows in $C$ such that $f=f'\circ a$ and $g=g'\circ a$.
This gives us a sense in which a universal object solves an optimization problem. Terminal and initial objects are the categorical analog of a minimum or maximum element in a poset (or preorder). Indeed a poset can be regarded as a category, and the category will have an initial element if and only if the poset has a maximal element.
Then since we get universal objects by constructing an auxiliary category and asking whether or not it has a terminal/initial object, we are very literally creating an optimization problem and asking if we can solve it. Now this leads to the natural question, what are we optimizing for? Why can we say it is the most efficient solution? Also as you ask in your question, why do we require a terminal object instead of just an object to which every other object has a morphism?
Second take (But why can we say that solving this optimization problem is finding the most efficient solution?)
Let's reconceptualize universal objects. Instead of thinking of universal objects as being initial or terminal objects in an auxiliary category, let's talk about information.
What information does an object in a category carry? Well, given some object $X\in C$, for any other object $Y$, we can construct the set $\newcommand\Hom{\operatorname{Hom}}\Hom(Y,X)$. Moreover for any morphism $f:Y\to Z$, we get a morphism $\Hom(Z,X)\to \Hom(Y,X)$ given by $g\mapsto g\circ f$. Thus to an object $X$ in $C$, we can associate a functor $h_X : C^\newcommand\op{\text{op}}\op\to\newcommand\Set{\mathbf{Set}}\Set$. It turns out that this functor encodes all of the (categorical) information present in the original object, in the sense that the functor determines $X$ (up to isomorphism). This is the content of the Yoneda lemma.
What does this have to do with universal objects, you might ask.
Well, suppose we have a functor $F:C^\op\to \Set$. This encodes some information about the category. Let's stick with products (though it may not be clear what the relation to products is yet). If $X$ and $Y$ are objects of $C$, then we could define $F(Z)$ to be the set of pairs of morphisms $(f,g)$ with $f:Z\to X$ and $g:Z\to Y$, and if $a:Z\to Z'$ is a morphism, we can define $F(a):F(Z')\to F(Z)$ by $F(a)(f',g') = (f'\circ a, g'\circ a)$.
Then $F(Z)$ encodes the information about pairs of morphisms from $Z$ to objects $X$ and $Y$. We might ask whether there is a single object in $C$ that encodes this same information.
What would this mean, well, it would mean that there is some object $X\times Y$ such that $h_{X\times Y}\simeq F$. In this case, we say $X\times Y$ represents the functor $F$. (There's actually a bit of extra data here, but we'll talk about that in the next section)
We've gotten fairly abstract, and apparently far from the context of the question. However, now we can talk about why we can say that this is the most efficient solution.
Suppose we just wanted to encode the information in $F$. Then all we need is an object $D$ such that we have a natural transformation $F\to h_D$, all of whose components are injections. Encoding it efficiently as possible should mean that $h_D$ doesn't carry any extra information, which corresponds to the natural transformation actually being a bijection.
Alternatively, by saying $h_D$ encodes all of the information of $F$, we could have the natural transformation go from $h_D$ to $F$, and require the components to be surjective, and then maximum efficiency is again when the components are bijections. It turns out that this second formulation is more useful, because of how the Yoneda lemma works.
Synthesis (What does representing a functor have to do with uniqueness of the morphism to the terminal object in the auxiliary category?)
In the last section I focused on representability of functors, which corresponds to the terminal object sort of universal object, so let's connect those two concepts.
There are two key points to connect the two viewpoints. The first is we should be a little more careful about a representing object.
It turns out that if $X$ represents $F$, then not only do we need $h_X\simeq F$, but the particular natural isomorphism is important too. Luckily, the Yoneda lemma tells us what the natural transformations from a hom functor to an arbitrary contravariant functor are. It turns out that $\Hom(h_X,F)\simeq F(X)$. Thus a natural isomorphism from $h_X$ to $F$ corresponds to a particular element of the set $F(X)$.
The second key point is to construct an auxiliary category that encodes the information of a functor, so that we can apply the perspective from the first section. Suppose then that we have some information encoded in a contravariant functor $F$. We can
construct an auxiliary category whose objects are pairs $(X,y)$ with $y\in F(X)$, and whose morphisms $g:(X,y)\to (X',y')$ are maps $g:X\to X'$ such that $F(g)(y')=y$.
By the Yoneda lemma, the objects $(X,y)$ of this category encode natural transformations $\psi_{X,y}$ from $h_X$ to $F$.
There is at least one morphism to $(X,y)$ from any other object $(X',y')$ in the auxiliary category if and only if the natural transformation $h_X\to F$ is surjective. (If you're not familiar with the Yoneda lemma, you'll probably want to skip or skim the proof.)
Proof.
The proof is just unwrapping the definitions.
First suppose that for any other object $(X',y')$, there is a map $g:(X',y')\to (X,y)$.
The natural transformation $\psi_{X,y,X'} : \Hom(X',X)=h_X(X')\to F(X')$ is defined by $g\mapsto F(g)(y)$. Let $y' \in F(X')$. Let $g : (X',y')\to (X,y)$, by the assumption. Then by definition of morphisms in the augmented category, $g\in \Hom(X',X)$ with $F(g)(y')=y$. Then $\psi_{X,y,X'}(g) = F(g)(y)=y'$. Hence $\psi_{X,y,X'}$ is surjective.
Now suppose conversely that $\psi_{X,y,X'}$ is surjective for all $X'$. Then for any $y'\in F(X')$, there is some $g\in h_X(X')=\Hom(X,X')$ such that $\psi_{X,y,X'}(g) = y'$.
However $\psi_{X,y,X'}(g) = F(g)(y)$. Thus $F(g)(y)=y'$, so $g$ is a morphism from $(X',y')$ to $(X,y)$, as desired. $\blacksquare$
The natural transformation $\psi_{X,y} : h_X\to F$ determined by the object $(X,y)$ in the auxiliary category is injective if and only if there is at most one morphism $g:(X',y')\to (X,y)$ for any other pair $(X',y')$.
Proof.
Once again, we are just unwrapping the definitions.
Suppose that there is at most one morphism to $(X,y)$ from any other pair. Then if $\psi_{X,y,X'}(g) =\psi_{X,y,X'}(h)$, we have that $F(g)(y)=F(h)(y)$, so $g$ and $h$ are both morphisms from $(X,F(g)(y))$ to $(X,y)$. By uniqueness, $g=h$. Therefore the components of the natural transformation are injective.
Conversely, suppose that $\psi_{X,y,X'}$ is injective for all $X'$. Then if $g$ and $h$ are both morphisms from $(X',y')$ to $(X,y)$, then
$$\psi_{X,y,X'}(g) = F(g)(y) = y' = F(h)(y) = \psi_{X,y,X'}(h),$$
so $g=h$. Thus morphisms to $(X,y)$ are unique. $\blacksquare$
Corollary. The natural transformation $\psi_{X,y}:h_X\to F$ associated to a pair $(X,y)$ in the auxiliary category is a natural isomorphism if and only if $(X,y)$ is terminal in the auxiliary category, and existence of the morphisms in the auxiliary category corresponds to the pair $(X,y)$ containing all the information of $F$ (in the sense of surjectivity of the components), and uniqueness of the morphisms is what corresponds to efficiency, (in the sense of injectivity of the components).
