Here is currently my favorite way to motivate forcing, and I conjecture that it works for most "real" mathematicians (non-logicians). A proof/disproof is left to the reader.

The forcing relation is indeed daunting at first sight; I was never able to remember the definitions until learning the Boolean-valued model approach. Meanwhile, I'm a material set theorist by training and don't want to go so far as advertising topos-theoretic forcing (at least before I understand how iterated forcing works in that setting), so let me advertise Boolean-valued model. Another reason I like the Boolean approach is the nice analogy to probability theory.

First, one doesn't need to know every axiom of $\mathsf{ZFC}$ in order to understand forcing, but several concepts especially helpful to be aware of are models, independence and absoluteness. Of course, models are everywhere: groups are models of group axioms, both $\mathbb{R}^2$ and Poincare disk are models of Hilbert's plane geometry axioms, etc. A model of set theory isn't any different: although we don't usually think of it this way, a model of $\mathsf{ZFC}$ is just a well-founded extensional directed graph (of course I am ignoring the subtleties around set model and class model) that satisfies some extra axioms. Independence is also everywhere: a group may or may not be abelian, a model of plane geometry may or may not satisfy Parallel Postulate, and it's easy to show the independence of power set axiom, replacement axiom, etc., from the rest of $\mathsf{ZFC}$. Absoluteness also has plenty of examples; if $A$ is an abelian group, $a\in A$ and $A$ satisfies the statement $\exists x\ x+x=a$ ($a$ can be divided by two in $A$), and $B$ is a subgroup, then $\exists x\ x+x=a$ isn't necessarily true in $B$. Nevertheless, bounded formulas and more generally $\Delta_1$ formulas are absolute between (transitive) models of set theory. This is because $\Delta_1$ formulas represent recursive constructions, and recursive constructions are intuitively absolute: think about $\Delta_1$ formulas in Peano arithmetic. In particular the constructible universe $L$ is absolute.

Now we can talk about forcing. Say we want to create a model of $\mathsf{ZFC}+\lnot\mathsf{CH}$. The method of inner model (like $L$) cannot possibly work, as observed by Shepherdson and Cohen, due to the existence of minimal model. So let's try the other way round: start with a model $M$ and expand it instead of shrinking it. For simple reasons we should not choose $M$ to be the whole universe $V$ or some level of von Neumann hiearachy $V_\kappa$, so maybe let's choose $M$ to be as small as possible, say countable, so that there are many things outside of $M$ that we can potentially throw into $M$.

Let $G\subseteq\omega$ be a set of natural numbers that is not in $M$; we want to adjoin it to $M$ and create a larger model $M[G]$ having the same ordinals. If we can add one then presumably we can add many, plus if we manage to add one then it already shows the independence of $V=L$ (by absoluteness of $L$ and the fact that $M[G]$ has the same ordinals), which is not bad. There are simple examples showing not all $G$ would work, but let's not worry about that yet, and think about what $M[G]$ should be. It is certainly not $M\cup\{G\}$ since the latter doesn't satisfy any interesting set theory. At the very least, $M[G]$ should contain all the sets "generated by $G$ over $M$ using simple operations", such as $\omega\setminus G$, $G\times G$, $\{n\in\omega: \text{the $n$-th prime is in }G\}$, etc. Note that:

$\omega\setminus G=\{n\in\omega:n\notin G\}$

$G\times G=\{(m,n)\in\omega\times\omega:m\in G\land n\in G\}$

$\{n\in\omega: \text{the $n$-th prime is in }G\}=\{n\in\omega:p_n\in G\}$, where $p_n$ denotes the $n$-th prime.

All these sets have the form $u=\{x\in X:b_x\}$; we can view it as a function $u$ that sends $x\in X$ to $b_x$, where $X$ is a set in $M$ and $b_x$ is a Boolean combination of statements of the form $n\in G$. I want to further rewrite these sets as follows. Let $\mathcal{G}$ be a fixed symbol. Consider the set $B$ of all Boolean combination of the expressions $n\in\mathcal{G}$, such as $(0\in\mathcal{G})\land(1\in\mathcal{G}\lor3\notin\mathcal{G})$. This is the free Boolean algebra with countably many generators $b_n$, where $b_n$ stands for $n\in\mathcal{G}$. A Boolean algebra is a structure $(B,\lor,\land,*,0,1)$ that behaves similar to union, intersection and complementation of sets; in our example, $b^*$ is the negation of $b$, e.g., $(0\in\mathcal{G})^*=0\notin\mathcal{G}$ and $[(0\in\mathcal{G})\land(1\in\mathcal{G}\lor3\notin\mathcal{G})]^*=[(0\notin\mathcal{G})\lor(1\notin\mathcal{G}\land3\in\mathcal{G})]$.

It should be emphasized that $\mathcal{G}$ is just a symbol intended to make $B$ more suggestive, in contrast to $G$, which is an actual subset of $\omega$. In particular $B\in M$. Any free Boolean algebra on countably many generators might be used as $B$ (as long as it is in $M$).

For a real number $G\subseteq\omega$, say that it satisfies $b\in B$ if $b$ is true if we plug $G$ into $\mathcal{G}$; for example, $G$ satisfies the statement $(0\in\mathcal{G})\land(1\in\mathcal{G}\lor3\notin\mathcal{G})$ iff $0\in G$ and at least one of $1\in G$ and $3\notin G$ happens. For a function $u:X\rightarrow B,x\mapsto b_x$, define the interpretation of $u$ under $G$ by $u_G=\{x:G\text{ satisfies } b_x\}$. Now observe that:

$\omega\setminus G=\{(n,b_n^*):n\in\omega\}_G$;

$G\times G=\{((m,n),b_m\land b_n):m,n\in\omega\}_G$;

$\{n\in\omega: \text{the $n$-th prime is in }G\}=\{(n,b_{p_n}):n\in\omega\}_G$.

So they are all of form $u_G$, where $u:X\rightarrow B$ is a function, and most importantly $X$ and $u$ are in $M$. This suggests that a set in $M[G]$, roughly speaking, consists of an $M$-part and a $G$-part. The $M$-part is a function $u:X\rightarrow B,x\mapsto b_x$; we can think of $u$ as a "random subset" of $X$, and $b_x$ as the "probability" of $x\in u$. And once we choose a point $G$ from the "sample space", the random set $u$ is determined to be $u_G$.

These random sets are more commonly called names: imagine that people living in $M$ cannot see $G$ or other sets in the extension $M[G]$, but nevertheless can name them and even reason about them. In particular there is a name for $G$, namely $\dot{G}=\{(n,b_n):n\in\omega\}$; it has the property that $\dot{G}_G=G$ regardless of $G$. If we let $\dot{G}'=\{(n,b_n^*):n\in\omega\}$, then people living in $M$ can see that "$\dot{G}'$ is the complement of $\dot{G}$", although they don't know any particular element of $G$.

Now comes one of the key ideas of forcing: pretend that we are the people living in $M$, don't know what $G$ is, but have names for $G$ and all the sets it generate. Although we don't know whether $3\in\dot{G}$ or not, we know its probability is $b_3$ or $3\in\mathcal{G}$. More generally, it turns out we can calculate the probability of any statement $\varphi$ about $M^B$, the collection of random sets in $M$; the probability will be a Boolean value, that is an element $||\varphi||\in B$. And the miracle is that every $\mathsf{ZFC}$ axiom holds with probability $1$ in this "probabilistic model" $M^B$. Incidentally, in this approach it's not important anymore that we start with a countable $M$---we could have started with the whole universe $V$ and form the probabilistic model $V^B$.

Unfortunately, our previous definition of random set, namely a function from some set $X$ to $B$, is problematic in two ways. First there are sets such as $\{n\in\omega:G\text{ contains some element divisible by }n\}$ that definitely should be in $M[G]$; what's the corresponding random set $u$? We would like to define $u(n)$ to be the ``sum'' $\displaystyle\sum_{n\mid m}b_m$, but by definition $B$ only contains finite combinations of the $b_n$s. If we replace $B$ by its Boolean completion, then there is a natural definition of the sum of an arbitrary set $A\subseteq B$: it's just the supremum $\bigvee A$. So let's assume $B$ is complete from now on.

Another issue is that sets belongt o other sets, so we should also allow random sets to belong to other random sets, in a random way. This naturally leads to the probabilistic von Neumann hierarchy: $V^B_0=\emptyset$, $V^B_{\alpha+1}$ is the set of functions from $V^B_{\alpha}$ to $B$ (it's actually a bit more convenient to take partial functions), and at limit ordinals take union. In one sentence, a random set is a random set of random sets. Every set $x\in V$ has a canonical name $\check{x}$ in $V^B$, defined recursively by letting $\check{y}$ belong to $\check{x}$ with probability $1$ for all $y\in x$.

The technical part is to actually define the probability, or Boolean value of an arbitrary statement $\varphi$ in $V^B$; this is the counterpart of the recursive definition of forcing relation in poset approach. The gist is really the case of atomic formulas $||u\in v||$ and $||u=v||$; the propositional connectives and quantifiers are easily handled, thanks to the completeness of $B$. I mentioned above that $u(x)$ can be viewed as the probability of $x\in u$; in fact this is true only for random sets that are simple enough, in general we should interpret $u(x)=b$ as $x\in u$ with probability at least $b$. For example, since $V^B$ is supposed to be an extension of $V$, it is reasonable to expect that for any canonical names $\check{x}$ and $\check{y}$, $||\check{x}=\check{y}||$ is $1$ if $x=y$ and $0$ if $x\neq y$, similarly for $||\check{x}\in\check{y}||$. If $u$ is a random set whose domain $\mathrm{dom}(u)$ is a set of canonical names, it is natural to let $||\check{x}\in u||$ be $u(\check{x})$ if $\check{x}\in\text{dom}(u)$, and $0$ otherwise. Now suppose $x,y,z$ are different sets in $V$ and let

$u=\{(\check{x},a),(\check{y},b)\}$,

$v=\{(\check{y},c),(\check{z},d)\}$,

what should $||u=v||$ be? Intuitively, $u=\{y\}$ with probability $a^*\land b$ and $v=\{y\}$ with probability $c\land d^*$, and that's the only way they could be equal, so the probability of $u=v$ is $a^*\land b\land c\land d^*$. Next consider

$w=\{(u,p),(v,q)\}$.

What is $||u\in w||$? It is certainly at least $p$, but also $u=v$ with probability $a^*\land b\land c\land d^*$ and $v\in w$ with probability at least $q$, and in order for $u=v\land v\in w\rightarrow u\in w$ to hold in our Boolean-valued model, $u\in w$ should have probability at least $a^*\land b\land c\land d^*\land q$. Altogether, $||u\in w||$ is at least $p\lor(a^*\land b\land c\land d^*\land q)$; there's no obvious reason it should be any bigger, so we make this the definition of $||u\in w||$. Formally, we define $||u=v||$ and $||u\in v||$ simultaneously by transfinite induction, following this line of thought.

Once we show that this definition indeed gives us a Boolean-valued model, namely it has properties such as $||u=v||\land||v=w||\leq||u=w||$, it's not difficult to verify the $\mathsf{ZFC}$ axioms all hold with probability $1$. For example, to construct the subset of $u$ consisting of elements with property $\varphi$, simply "reweight" elements of $u$ according to their probability of satisfying $\varphi$. There is no need to pass to countable model: one can directly argue with the Boolean-valued model $V^B$ to do independence proof, using the Boolean version of soundness theorem. If you insist, though, it is not difficult at all to relativize all of the above to some countable model $M$, choose a generic ultrafilter $G$ of the Boolean algebra $B$ and get a good old generic extension $M[G]$. The proof of the truth and definability lemmas are also cleaner and somewhat motivated: in order that $||\varphi||\in G$ iff $M[G]\models\varphi$, one naturally wants the filter $G$ to be generic for arguments to go through.

No intent to trivialize Cohen's accomplishments, but (something like) Boolean-valued models were long considered before him, although nobody came up with the idea of using them to prove independence result; see Dana Scott's preface to the book Boolean-Valued Models and Independence Proofs. Another interesting sentence from The Origins of Forcing, an interview of Cohen by Gregory H. Moore:

Cohen recalls that in the eyes of various logicians his forcing results went from being incorrect, to being extremely difficult to understand, to being easy, and finally to being already present in the literature.

I believe this shows that to some extent, Boolean-valued model does make forcing easier to understand.