I think maybe the best way to explain the notion of likelihood is to consider a concrete example. Suppose I have a sample of IID observations drawn from a Bernoulli distribution with unknown probability of success $p$: $X_i \sim {\rm Bernoulli}(p)$, $i = 1, \ldots, n$, so the joint probability mass function of the sample is $$\Pr[{\boldsymbol X} = \boldsymbol x \mid p] = \prod_{i=1}^n p^{x_i} (1-p)^{1-x_i}.$$ This expression also characterizes the likelihood of $p$, given an observed sample $\boldsymbol x = (x_1, \ldots, x_n)$: $$L(p \mid \boldsymbol x) = \prod_{i=1}^n p^{x_i} (1-p)^{1-x_i}.$$ But if we think of $p$ as a random variable, this likelihood is not a density: $$\int_{p=0}^1 L(p \mid \boldsymbol x) \, dp \ne 1.$$ It is, however, proportional to a probability density, which is why we say it is a likelihood of $p$ being a particular value given the sample--it represents, in some sense, the relative plausibility of $p$ being some value for the observations we made.
For instance, suppose $n = 5$ and the sample was $\boldsymbol x = (1, 1, 0, 1, 1)$. Intuitively we would conclude that $p$ is more likely to be closer to $1$ than to $0$, because we observed more ones. Indeed, we have $$L(p \mid \boldsymbol x) = p^4 (1 - p).$$ If we plot this function on $p \in [0,1]$, we can see how the likelihood confirms our intuition. Of course, we do not know the true value of $p$--it could have been $p = 0.25$ rather than $p = 0.8$, but the likelihood function tells us that the former is much less likely than the latter. But if we want to determine a probability that $p$ lies in a certain interval, we have to normalize the likelihood: since $\int_{p=0}^1 p^4(1-p) \, dp = \frac{1}{30}$, it follows that in order to get a posterior density for $p$, we must multiply by $30$: $$f_p(p \mid \boldsymbol x) = 30p^4(1-p).$$ In fact, this posterior is a beta distribution with parameters $a = 5, b = 2$. Now the areas under the density correspond to probabilities.
So, what we have essentially done here is applied Bayes' rule: $$f_{\boldsymbol \Theta}(\boldsymbol \theta \mid \boldsymbol x) = \frac{f_{\boldsymbol X}(\boldsymbol x \mid \boldsymbol \theta) f_{\boldsymbol \Theta}(\boldsymbol \theta)}{f_{\boldsymbol X}(\boldsymbol x)}.$$ Here, $f_{\boldsymbol \Theta}(\boldsymbol \theta)$ is a prior distribution on the parameter(s) $\boldsymbol \theta$, the numerator is the likelihood $L(\boldsymbol \theta \mid \boldsymbol x) = f_{\boldsymbol X}(\boldsymbol x \mid \boldsymbol \theta) f_{\boldsymbol \Theta}(\boldsymbol \theta) = f_{\boldsymbol X, \boldsymbol \Theta}(\boldsymbol x, \boldsymbol \theta)$ which is also the joint distribution of $\boldsymbol X, \boldsymbol \Theta$, and the denominator is the marginal (unconditional) density of $\boldsymbol X$, obtained by integrating the joint distribution with respect to $\boldsymbol \theta$ to find the normalizing constant that makes the likelihood a probability density with respect to the parameter(s). In our numerical example, we implicitly took the prior for $f_{\boldsymbol \Theta}$ to be uniform on $[0,1]$. It can be shown that, for a Bernoulli sample, if the prior is ${\rm Beta}(a,b)$, the posterior for $f_{\boldsymbol \Theta}$ is also Beta, but with parameters $a^* = a+\sum x_i$, $b^* = b + n - \sum x_i$. We call such a prior conjugate (and refer to this as a Bernoulli-Beta conjugate pair).
