Let $\mu$ be a finite, strictly positive measure on $\mathbb{R}$, and let $k$ be a measurable positive-definite kernel. Assume $k$ is bounded, and let $T:L^2(\mu)\rightarrow L^2(\mu)$ be defined by $$ Tf(x):=\int k(x,y) f(y)d\mu(y).$$
Does the representation $$k(x, y) = \sum_{j=1}^\infty \lambda_j \varphi_j(x) \varphi_j(y)$$ hold uniformly on $(x,y)$?
Here, the $(\lambda_j, \varphi_j)$ pairs are defined via $$T\varphi_j = \lambda_j\varphi_j.$$
I'm assuming that the functions $\varphi_j$ are real-valued. The Cauchy-Schwartz inequality on $\ell^2(\mathbb{N})$ gives
\begin{align*}\sum_{j=1}^{\infty} |\lambda_j \varphi_{j}(x) \varphi_{j}(y)| &= \sum_{j=1}^{\infty} |\sqrt{\lambda_j} \varphi_{j}(x) \sqrt{\lambda_j}\varphi_{j}(y)| \leq \sum_{j=1}^{\infty} \lambda_j \varphi_j(x)^2 \sum_{j=1}^{\infty}\lambda_j \varphi_j(y)^2 \\ &= k(x,x) k(y,y) \leq C^2\end{align*}
As $k(x,y)$ is bounded. This shows uniform convergence, assuming you know already that the equality holds pointwise.
EDIT: I made an implicit assumption in the following argument that I just became aware of now. Let's assume that the $\varphi_j$ are continuous (that would follow if we would assume $k$ to be continuous in both variables). Then we know the equality holds pointwise for some subsequence $\sum_{j=1}^{n_k} \lambda_j \varphi_j(x) \varphi_j(y)$. Applying the argument above shows uniform convergence for this subsequence. This implies that there is a $k_0$ such that $$\sum_{j=n_k}^{\infty} \|\lambda_j \varphi_j(.) \varphi_j(.)\|_{\infty} < \epsilon, \quad k\geq k_0$$ which implies the existence of a $N$ such that
$$ \sum_{j=N}^{\infty} \|\lambda_j \varphi_j(.) \varphi_j(.)\|_{\infty} < \epsilon, \quad n\geq N$$
Now, if the $\varphi_j$ are continuous, we can identify $(\sum_{j=1}^{n}\lambda_j \varphi_j(.)\varphi_j(.))_{n\in\mathbb{N}}$ as a Cauchy-sequence in $C([0,1]\times[0,1])$ with the supremum norm. As this is a complete space, uniform convergence follows.