This notion of convergence is not often refered to; I think mostly because it does not come from a topology. But this an excellent notion of convergence, probably the best we can put on the space of (continuous) functions. I can tell you a few things about it - but only in the case where all the functions are continuous:
First observation: This notion of convergence does not depend on the uniform structure of $Y$, and we don't even need this uniform structure to define it:
One will say that a net $f_{\alpha}$ of functions $X \rightarrow Y$ tends to a function $f:X\rightarrow Y$ if:
For all $U$ open subsets of $Y$, $f^{-1}(U)$ admit a covering by open subsets $V_i$ such that for each $i$ there exists a $\alpha_i$ such that for all $\alpha \geqslant \alpha_i$ one has $f_{\alpha}(V_i) \subset U$.
Proposition: (if $f$ is continuous) This notion of convergence is equivalent to yours.
Prof: Assume convergence in my definition, then take $x_0 \in X$ and $\epsilon >0$. Let $U=\{ y | d(f(x_0),y)<\epsilon \}$ there exists a covering of $f^{-1}(U)$ by $V_i$ such that... In particular, $x_0$ is in one of the $V_i$, and for each $x \in V_i$ one has, for all $\alpha \geqslant \alpha_i$, $d(f_{\alpha}(x),f(x_0)) <\epsilon$ which is not exactly your definition, but you can restrict $V_i$ a little more using the continuity of $f$ such that additionally for all $x\in V_i$, $d(f(x),f(x_0))<\epsilon$ and we are done.
The converse is essentially the same: If $ f_{\alpha}$ converge in your sense, then for each $U \subset Y$, for each $x \in f^{-1}(U)$ there exist an $\epsilon$ such that the $\epsilon$ neighbourhood of $f(x)$ is in $U$ and applying both your definition and the continuity of $f$ one can find a "$V_i$" containing $x$.
Second observation Either with your definition or mine, The space of continuous functions is closed for this notion of convergence. Also, the evaluation map: $X \times Y^X \rightarrow Y$ is jointly continuous (on continuous function of course). In particular, when this notion of convergence on continuous function comes from a topology, then this topology on $Y^X$ is the exponential in the category of topological space. As it is well know that the category of topological does not admit all exponentials, this notion of convergence will not always correspond to a topology (essentially, it will fail as soon as $X$ is not locally compact and $Y$ not too trivial).
In fact if you denote $K$ to one point compactification of $\mathbb{N}$, then a sequence $f_n$ tend to $f$ for this notion of convergence exactly if the corresponding map $K \rightarrow Y^X$ is continuous in the "exponential" sense, i.e. if $ K \times X \rightarrow Y$ is continuous.
Thrid observation: This will probably not speak to everyone, but this notion as a really nice interpretation in topos theory and more precisely in the philosophy of geometric logic. A function from $X$ to $Y$ can be thought of as a "point" of the topos $sh(Y)$ in the internal logic of the topos $sh(X)$. This notion of convergence is exactly the notion of convergence of points of $sh(Y)$ interpreted in the internal logic of $sh(X)$.
Fourth observation: This notion of convergence (with my definition) actually has a name (althoug not widely used) it is "continuous convergence" which is defined on arbitrary convergence space by: a sequence $f_{\alpha}$ converge to $f$ if for each net $x_{\beta}$ converging to $x$ the net $f_{\alpha}(x_{\beta})$ converge to $f(x)$. There is a few paper about this (a part of which are in German...) I think This paper is a good starting point (It seems freely available and there is more reference in its bibliography) but I'm not really familiar with this literature.
Finally (but this might be subjective) I'm afraid that for non continuous functions this notion does not make sense so well. I mean, you can still use the definition of course, but the proof of the equivalence of the various possible definition of this notion of convergence as well as the various theoretical interpretation that we can give seems to rely a lot on continuity...