I think this is largely unnecessary splitting hairs.

Conditional probability $P(x|y)\equiv P(X=x|Y=y)$ of $x$ given $y$ is defined for two random variables $X$ and $Y$ taking values $x$ and $y$. But we can also talk about probability $P(x|\theta)$ of $x$ given $\theta$ where $\theta$ is not a random variable but a parameter.

Note that in both cases the same term "given" and the same notation $P(\cdot|\cdot)$ can be used. There is no need to invent different notations. **Moreover, what is called "parameter" and what is called "random variable" can depend on your philosophy,** but the math does not change.

The first quote from Wikipedia states that $\mathcal{L}(\theta |x) = P(x | \theta)$ by definition. Here it is assumed that $\theta$ is a parameter. The second quote says that $\mathcal{L}(\theta |x)$ is *not* a conditional probability. This means that it is not a conditional probability of $\theta$ given $x$; and indeed it cannot be, because $\theta$ is assumed to be a parameter here.

In the context of Bayes theorem $$P(a|b)=\frac{P(b|a)P(a)}{P(b)},$$ both $a$ and $b$ are random variables. But we can still call $P(b|a)$ "likelihood" (of $a$), and now it is also a *bona fide* conditional  probability (of $b$). This terminology is standard in Bayesian statistics. Nobody says it is something "similar" to the likelihood; people simply call it the likelihood.

**Note 1:** In the last paragraph, $P(b|a)$ is obviously a conditional probability of $b$. As a likelihood $\mathcal L(a|b)$ it is seen as a function of $a$; but it is not a probability distribution (or conditional probability) of $a$! Its integral over $a$ does not necessarily equal $1$. (Whereas its integral over $b$ does.)

**Note 2:** Sometimes likelihood is defined up to an arbitrary proportionality constant, as emphasized by @MichaelLew (because most of the time people are interested in likelihood *ratios*). This can be useful, but is not always done and is not essential.

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See also http://stats.stackexchange.com/questions/2641/ and in particular @whuber's answer there.

I fully agree with @Tim's answer in this thread too (+1).