= Solution
Let $m=\dim M$ and $q=m-\dim Y$. Here a closed submanifold means compact and without boundary, which is the condition needed for a compactly supported dual class. The orientations of $M$ and $Y$ orient the <normal bundle> $\nu_Y$. Its <Thom class> is the unique relative class of degree $q$ restricting to the chosen orientation generator on every normal fiber. A <tubular neighborhood> identifies this class with one near $Y$ in $M$. Choose a compact normal disc neighborhood and extend the relative class to <compactly supported cohomology> of $M$. Equivalently, apply <Poincare duality> $H_c^q(M;\mathbb Q)\cong H_{m-q}(M;\mathbb Q)$ to the pushed-forward <fundamental class> of $Y$:
$$
\boxed{\varepsilon_Y=\operatorname{PD}_M(i_*[Y])\in H_c^q(M;\mathbb Q).}
$$
This <compactly supported dual of a compact submanifold> is independent of the tubular neighborhood and represents intersection with $Y$. Fix the convention $\langle\varepsilon_Y\smile a,[M]\rangle=\langle i^*a,[Y]\rangle$ in complementary degree. Its image in ordinary <cohomology> restricts to zero on $M\setminus Y$, because it comes from relative <cohomology> supported near $Y$. If a submanifold is only closed as a subset but is noncompact, its usual dual lies in ordinary <cohomology> instead; compact support is not asserted in that situation.
Now let $\Delta\subset N\times N$ be the diagonal, with the orientation inherited from $N$, and let $\delta=\varepsilon_\Delta\in H^n(N\times N;\mathbb Q)$. For each $i$, choose a basis $a_{i,j}$ of $H^i(N;\mathbb Q)$ and its dual basis $b_{i,j}$ of $H^{n-i}(N;\mathbb Q)$ satisfying $\langle a_{i,j}\smile b_{i,k},[N]\rangle=\delta_{jk}$. The <Poincare duality pairing> and the <Künneth theorem> give the <cohomology class of the diagonal>
$$
\delta=\sum_{i,j}(-1)^{i(n-i+1)}p_1^*b_{i,j}\smile p_2^*a_{i,j}.
$$
The sign can be checked from the defining identity
$$
\left\langle\delta\smile(p_1^*u\smile p_2^*v),[N\times N]\right\rangle=\langle u\smile v,[N]\rangle.
$$
For $|u|=i$, moving the degree-$i$ class in the second factor past $u$ contributes $(-1)^i$, and interchanging $b_{i,j}$ and $u$ contributes $(-1)^{i(n-i)}$; the displayed coefficient cancels their product. Equivalently its exponent is $ni$ modulo two.
The <cohomological pushforward between closed oriented manifolds> is characterized by
$$
\langle f^!c\smile b,[N]\rangle=\langle c\smile f^*b,[M]\rangle,
$$
for complementary degrees. This follows immediately from its definition by <Poincare duality> and naturality of evaluation on pushed-forward homology. In degree $i$, the <matrix trace> of $f^!g^*$ on $H^i(N;\mathbb Q)$ is consequently
$$
\operatorname{tr}(f^!g^*)=\sum_j\langle g^*a_{i,j}\smile f^*b_{i,j},[M]\rangle.
$$
For finite-dimensional maps between two possibly different vector spaces, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. Thus this <matrix trace> equals that of $g^*f^!$ on $H^i(M;\mathbb Q)$. Pulling back the diagonal formula by $F=(f,g)$ and applying <graded commutativity of the cup product> now yields
$$
\boxed{\left\langle(f,g)^*\delta,[M]\right\rangle=\sum_i(-1)^i\operatorname{tr}(g^*f^!)=L(f,g).}
$$
This establishes the <Lefschetz coincidence number> with the requested sign convention.
If $f(m)\ne g(m)$ for every $m$, the map $(f,g)$ factors through $(N\times N)\setminus\Delta$. The diagonal class restricts to zero there, so its pullback is zero and $L(f,g)=0$. Taking the contrapositive proves the <Lefschetz coincidence theorem>:
$$
\boxed{L(f,g)\ne0\quad\Longrightarrow\quad f(m)=g(m)\text{ for some }m\in M.}
$$
No transversality or isolated-coincidence assumption is needed. This cohomological argument also applies to continuous maps, which covers the last request.
For a map $f:M\to N$ of degree $d$, the adjoint identity above gives, for all complementary classes $a,b$ on $N$,
$$
\langle f^!f^*a\smile b,[N]\rangle=\langle f^*(a\smile b),[M]\rangle=d\langle a\smile b,[N]\rangle.
$$
Nondegeneracy therefore gives $f^!f^*=d\,\mathrm{id}$. Cyclic invariance of <matrix trace> proves the <self-coincidence number of a map of nonzero degree> formula
$$
L(f,f)=d\,\chi(N).
$$
In particular, $\mathbb{CP}^{2k}$ has one even-dimensional <cohomology> generator in each degree $0,2,\ldots,4k$, so $\chi(\mathbb{CP}^{2k})=2k+1$. Hence
$$
\boxed{L(f,f)=(2k+1)d\ne0.}
$$
If $g$ is homotopic to $f$, <homotopy invariance of cohomology> gives $g^*=f^*$ and thus $L(f,g)=L(f,f)$. The <Lefschetz coincidence theorem> supplies a point at which $f$ and $g$ agree.
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